mahs-dv algebra 1-2 q4 · a quadratic function is a function of the form f(x) = ax2 +bx+c where a;b...
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MAHS-DV Algebra 1-2 Q4
Adrienne Wooten
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Printed: March 5, 2014
AUTHORAdrienne Wooten
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Contents www.ck12.org
Contents
1 Quadratics 11.1 Graphs of Basic Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Transformations of Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Vertex Form of a Quadratic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.4 Graphs to Solve Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5 Use Square Roots to Solve Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 381.6 Solving Quadratics using Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.7 Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491.8 Solving Quadratic Equations by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . 551.9 Solutions Using the Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 671.10 Modeling with Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721.11 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
2 Non-Linear Relationships 792.1 Piecewise Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 802.2 Graphing Basic Absolute Value Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 842.3 Using the General Absolute Value Equation and the Graphing Calculator . . . . . . . . . . . . . 912.4 Graphs of Square Root Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 962.5 Shifts of Square Root Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022.6 Graphing Cubed Root Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
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CHAPTER 1 QuadraticsChapter Outline
1.1 GRAPHS OF BASIC QUADRATIC FUNCTIONS
1.2 TRANSFORMATIONS OF QUADRATIC FUNCTIONS
1.3 VERTEX FORM OF A QUADRATIC FUNCTION
1.4 GRAPHS TO SOLVE QUADRATIC EQUATIONS
1.5 USE SQUARE ROOTS TO SOLVE QUADRATIC EQUATIONS
1.6 SOLVING QUADRATICS USING FACTORING
1.7 COMPLETING THE SQUARE
1.8 SOLVING QUADRATIC EQUATIONS BY THE QUADRATIC FORMULA
1.9 SOLUTIONS USING THE DISCRIMINANT
1.10 MODELING WITH QUADRATIC FUNCTIONS
1.11 REFERENCES
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1.1. Graphs of Basic Quadratic Functions www.ck12.org
1.1 Graphs of Basic Quadratic Functions
Here you’ll learn how to graph and analyze the quadratic function, y = x2.
Look at the graph below. Does the graph represent a function? Do you know the name of the graph? Do you knowwhat makes the green point special? Do you notice any symmetry in the graph? Can you state the domain and rangefor the relation?
Watch This
Khan Academy Quadratic Functions1
MEDIAClick image to the left for more content.
Guidance
Until now you have been dealing with linear functions. The highest exponent of the independent variable (x) hasbeen one and the graphs have been straight lines. Here you will be learning about quadratic functions. A quadratic
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function is one of the form f (x) = ax2 +bx+ c where a,b and c are real numbers and a 6= 0. The highest exponentof the independent variable is two. When graphed, a quadratic function creates a parabola that looks like this:
or like this:
You can create your own graph by plotting the points created from a table of values. The most basic quadraticfunction is f (x) = x2. It is also called the parent function.
A parabola has a turning point known as the vertex. The vertex is the minimum value of the parabola if it opensupward and the maximum value if the parabola opens downward. When the graph opens downward, the y-valuesin the base table change to negative values. The basic quadratic function that opens downward has the equationf (x) =−x2. This is also called the end behavior of the graph. Given a quadratic function in the form of f (x) =ax2 +bx+ c (or f (x) = a(x−h)2 + k), the quadratic function is said to open up if a > 0 and open down if a < 0.
If a > 0, then the function has a minimum at the x-coordinate of the vertex. This means that the function isdecreasing for x-values less than (or to the left of) the vertex, and the function is increasing for x-values greater than(or to the right of) the vertex.
If a < 0, then the function has a maximum at the x-coordinate of the vertex. This means that the function isincreasing for x-values less than (or to the left of) the vertex, and the function is decreasing for x-values greater than(or to the right of) the vertex.
All parabolas have an axis of symmetry. The axis of symmetry is the vertical line that passes through the vertex ofthe parabola. The equation for the axis of symmetry is always the x−coordinate of the vertex. The equation for thisline is x =− b
2a
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1.1. Graphs of Basic Quadratic Functions www.ck12.org
Example A
For the basic quadratic function f (x) = x2, complete a table such that {x|−3≤ x≤ 3,x ε Z}. Z is the special symbolfor Integers.
Solution:
To complete the table of values, substitute the given x-values into the function f (x)= x2. If you are using a calculator,insert all numbers, especially negative numbers, inside parenthesis before squaring them. The operation that needsto be done is (−3)(−3) NOT −(3)(3).
f (x) = x2 f (x) = x2 f (x) = x2 f (x) = x2
f (x) = (−3)2 f (x) = (−2)2 f (x) = (−1)2 f (x) = (0)2
f (x) = 9 f (x) = 4 f (x) = 1 f (x) = 0
f (x) = x2 f (x) = x2 f (x) = x2
f (x) = (1)2 f (x) = (2)2 f (x) = (3)2
f (x) = 1 f (x) = 4 f (x) = 9
TABLE 1.1:
X f(x)−3 9−2 4−1 10 01 12 43 9
Example B
On a Cartesian plane, plot the points from the table for y = x2.
Solution:
The plotted points cannot be joined to form a straight line. To join the points, begin with the point (-3, 9) or the point(3, 9) and without lifting your pencil, draw a smooth curve. The image should look like the following graph.
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The arrows indicate the direction of the pencil as the points are joined. If the pencil is not moved off the paper,the temptation to join the points with a series of straight lines will be decreased. The points must be joined witha smooth curve that does not extend below the lowest point of the graph. In the above graph, the curve cannot gobelow the point (0, 0).
Example C
What are some unique characteristics of the graph of y = x2?
Solution:
1. The green point is located at the lowest point on the image. The curve does not go below this point.2. Every red point on the left side of the image has a corresponding blue point on the right side of the image.3. If the image was folded left to right along the y-axis that passes through the green point, each red point would
land on each corresponding blue point.4. The sides of the image extend upward.5. The red and the blue points are plotted to the right and to the left of the green point. The points are plotted left
and right one and up one; left and right two and up four, left and right 3 and up nine.
Example D
Find the vertex, the equation of the axis of symmetry, the y-intercept and the maximum or minimum of the function.
y = 2x2 +4x−3
To find the axis of symmetry, we use x =− b2a
x = 42(2)
x =−1 which is the axis of symmetry.
To find the vertex, use the value you found for the axis of symmetry as the x-coordinate of the vertex. To find they-coordinate, substitute that value of x in the original equation.
y = 2(−1)2 +4(−1)−3
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1.1. Graphs of Basic Quadratic Functions www.ck12.org
y =−5
The vertex is (−1,−5). Since a > 0 the parabola opens upward and has a minimum. The minimum is the y-coordinate of the vertex which in this case is −5.
The y-intercept always occurs at (0,c). So the y-intercept is (0,−3). All of this information can be used to graphthe function as seen in the figure below.
Concept Problem Revisited
The green point is the lowest point on the curve. The smooth curve is called a parabola and it is the image producedwhen the basic quadratic function is plotted on a Cartesian grid. The green point is known as the vertex of theparabola. The vertex is the turning point of the graph.
For the graph of y = x2, the vertex is (0, 0) and the parabola has a minimum value of zero which is indicated by they-value of the vertex. The parabola opens upward since the y-values in the table of values are 0, 1, 4 and 9. They-axis for this graph is actually the axis of symmetry. The axis of symmetry is the vertical line that passes throughthe vertex of the parabola. The parabola is symmetrical about this line. The equation for this axis of symmetry isx = 0. If the parabola were to open downward, the vertex would be the highest point of the graph. Therefore theimage would have a maximum value of zero.
The domain for all parabolas is D = {x|x ε R} or (−∞,∞) The range for the above parabola is R = {y|y≥ 0,y ε R}or [0,∞).
Vocabulary
Axis of SymmetryThe axis of symmetry of a parabola is a vertical line that passes through the vertex of the parabola. Theparabola is symmetrical about this line. The axis of symmetry has the equation x = the x−coordinate of thevertex.
ParabolaA parabola is the smooth curve that results from graphing a quadratic function of the form f (x) = ax2+bx+c.The curve resembles a U-shape.
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www.ck12.org Chapter 1. Quadratics
Quadratic FunctionA quadratic function is a function of the form f (x) = ax2 + bx+ c where a,b and c are real numbers anda 6= 0.
VertexThe vertex of a parabola is the point around which the parabola turns. The vertex is the maximum point of aparabola that opens downward and the minimum point of a parabola that opens upward.
Guided Practice
1. If the graph of y = x2 opens downward, what changes would exist in the base table of values?
2. If the graph of y = x2 opens downward, what changes would exist in the basic quadratic function?
3. Draw the image of the basic quadratic function that opens downward. State the domain and range for this parabola.
Answers:
1. If the parabola were to open downward, the x-values would not change. The y-values would become negativevalues. The points would be plotted from the vertex as: right and left one and down one; right and left two and downfour; right and left three and down nine. The table of values would be
TABLE 1.2:
X Y−3 −9−2 −4−1 −10 01 −12 −43 −9
2. To match the table of values, the basic quadratic function would have to be written as y =−x2.
3.
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1.1. Graphs of Basic Quadratic Functions www.ck12.org
The domain is D = {x|x ε N}. The range for this parabola is R = {y|y≤ 0,y ε N}.
Practice
Complete the following statements in the space provided.
1. The name given to the graph of f (x) = x2 is ____________________.2. The domain of the graph of f (x) = x2 is ____________________.3. If the vertex of a parabola was (-3, 5), the equation of the axis of symmetry would be ____________________.4. A parabola has a maximum value when it opens ____________________.5. The point (-2, 4) on the graph of f (x) = x2 has a corresponding point at ____________________.6. The range of the graph of f (x) =−x2 is ____________________.7. If the table of values for the basic quadratic function included 4 and -4 as x-values, the y-value(s) would be
____________________.8. The vertical line that passes through the vertex of a parabola is called ____________________.9. A minimum value exists when a parabola opens ____________________.
10. The turning point of the graph of f (x) = x2 is called the ____________________.11. For what values of x is f (x) = x2 increasing? For what values is it decreasing?12. Find the vertex, the equation of the axis of symmetry, the y-intercept and the maximum or minimum of the
following:
a. y =−3x2 +6x−1
b. y =−x2 +2x+1
c. y = x2−4x+5
d. y = 4x2−8x+9
13. A juggker is tossing a ball into the air. The height of the ball in feet can be modeled by the equationy =−16x2 +16x+5, where y represents the height of the ball in x seconds.
a. Graph the equation.
b. At what height is the ball thrown?
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1.2. Transformations of Quadratic Functions www.ck12.org
1.2 Transformations of Quadratic Functions
Here you’ll learn how to transform the basic quadratic functions (y = x2 and y = −x2) to make new quadraticfunctions.
Look at the parabola below. How is this parabola different from y = −x2? What do you think the equation of thisparabola is?
Watch This
Khan Academy Graphing a Quadratic Function
MEDIAClick image to the left for more content.
Guidance
Vertical Translations
Consider the parent function f (x) = x2 of the parabola as shown below.
Using a table of values or your graphing calculator, graph g(x) = x2 +2.
How is this graph related to the parent function f (x) = x2?
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www.ck12.org Chapter 1. Quadratics
Now graph g(x) = x2−2.
How is the graph of g(x) = x2 + k related to the parent function?
Functions in the form g(x) = x2+k shift up k units for k > 0 and shift down k units for k < 0 from the parent functionas shown below. This is called a vertical translation.
Horizontal Translations
Using a table of values or your graphing calculator, graph g(x) = (x−1)2.
How is this graph related to the parent function f (x) = x2?
Now graph g(x) = (x+1)2.
How is the graph of g(x) = (x−h)2 related to the parent function?
Functions in the form g(x) = (x− h)2 shift left h units for h > 0 and shift right h units for h < 0 from the parentfunction as shown below. This is called a horizontal translation.
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1.2. Transformations of Quadratic Functions www.ck12.org
Writing Equations for Quadratic Functions
Compare the function below to the graph of the parent function f (x) = x2.
Consider the translation of the parent function:
Horizontal translation of 2 units to the right.
Vertical translation of 1 unit up.
This will help us determine h and k for the function g(x) = (x−h)2 + k.
• h is the number of units the parent function is translated horizontally. If the function shifts to the right, h willbe positive; for a shift to the left, h will be negative.
• k is the number of units the parent function is translated vertically. If the function shifts up, k will be positive;for a shift down, k will be negative.
So, in our function, h = 2 and k = 1. If we plug those values in the equation above, we have g(x) = (x−2)2 +1.
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www.ck12.org Chapter 1. Quadratics
Vertical Stretches and Compressions
If a > 1, the parabola will be stretched vertically. If 0 < a < 1, the graph will be compressed vertically(sometimes called a vertical shrink). In the following figure, the parent function has been stretched g(x) = 2x2 andcompressed g(x) = 1
2 x2 vertically.
If a < 0, the graphs will be reflected across the x-axis (see below). This is called a vertical reflection which canalso be represented by − f (x). The graph of f (−x) is the reflection of the graph of across the y-axis but this resultsin the same function because f (−x) = (−x)2 = x2.
Writing the Equation for a Quadratic Function
Consider the graph below. How can we use the point (2,−1) to write the equation of the function?
A function whose graph is a parabola with vertex (0,0) always has the form f (x) = ax2. To write the rule for thefunction, we simply substitute the ordered pair (2,−1) into the equation and solve for a.
y = ax2
−1 = a(2)2
−1 = a(4)
a =−14 so our function is f (x) =−1
4 x2.
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1.2. Transformations of Quadratic Functions www.ck12.org
Putting It All Together
How can we graph g(x) = a(x− h)2 + k using the parent function f (x) = x2? All of the previous transformationscan be put together as a combination of transformations.
The sequence of graphs below shows how you can obtain the graph of g(x) =−2(x−2)2 +1
1. Start with the graph of y = x2.
2. Stretch the graph vertically by a factor of 2 to obtain the graph of y = 2x2.
3. Reflect the graph across the x-axis to obtain the graph of y =−2x2.
4. Translate the graph of y =−2x2 right 2 units and up one unit to obtain the graph of y =−2(x−2)2 +1.
Concept Problem Revisited
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This is the graph of y =−12 x2. The points are plotted from the vertex as right and left one and down one-half, right
and left 2 and down two, right and left three and down four and one-half. The original y-values of 1, 4 and 9 havebeen multiplied by one-half and then were made negative because the graph was opening downward. When they-values become negative, the direction of the opening is changed from upward to downward. This transformationis known as a vertical reflection. The graph is reflected across the x-axis.
Vocabulary
Horizontal translationThe horizontal translation is the change in the base graph y = x2 that shifts the graph right or left. It changesthe x−coordinate of the vertex.
TransformationA transformation is any change in the base graph y = x2. The transformations that apply to the parabola area horizontal translation, a vertical translation, a vertical stretch and a vertical reflection.
Vertical ReflectionThe vertical reflection is the reflection of the image graph in the x-axis. The graph opens downward and they-values are negative values.
Vertical StretchThe vertical stretch is the change made to the base function y = x2 by stretching (or compressing) the graphvertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the originalbase graph of y = x2.
Vertical TranslationThe vertical translation is the change in the base graph y = x2 that shifts the graph up or down. It changes they−coordinate of the vertex.
Guided Practice
1. Use the following tables of values and identify the transformations of the base graph y = x2.
X −3 −2 −1 0 1 2 3
Y 9 4 1 0 1 4 9
X −4 −3 −2 −1 0 1 2
Y 15 5 −1 −3 −1 5 15
2. Identify the transformations of the base graph y = x2.
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1.2. Transformations of Quadratic Functions www.ck12.org
3. Draw the image graph of y = x2 that has undergone a vertical reflection, a vertical stretch by a factor of 12 , a
vertical translation up 2 units, and a horizontal translation left 3 units.
Answers:
1. To identify the transformations from the tables of values, determine how the table of values for y = x2 compare tothe table of values for the new image graph.
• The x-values have moved one place to the left. This means that the graph has undergone a horizontal translationof –1.
• The y−coordinate of the vertex is –3. This means that the graph has undergone a vertical translation of –3. Thevertex is easy to pick out from the tables since it is the point around which the corresponding points appear.
• The points from the vertex are plotted left and right one and up two, left and right two and up eight. Thismeans that the base graph has undergone a vertical stretch of 2.
• The y-values move upward so the parabola will open upward. Therefore the image is not a vertical reflection.
2. The vertex is (1, 6). The base graph has undergone a horizontal translation of +1 and a vertical translation of+6. The parabola opens downward, so the graph is a vertical reflection. The points have been plotted such thatthe y-values of 1 and 4 are now 2 and 8. It is not unusual for a parabola to be plotted with five points rather thanseven. The reason for this is the vertical stretch often multiplies the y-values such that they are difficult to graph ona Cartesian grid. If all the points are to be plotted, a different scale must be used for the y-axis.
3. The vertex given by the horizontal and vertical translations and is (–3, 2). The y-values of 1, 4 and 9 must bemultiplied by 1
2 to create values of 12 ,2 and 4 1
2 . The graph is a vertical reflection which means the graph opensdownward and the y-values become negative.
Practice
The following table represents transformations to the base graph y = x2. Draw an image graph for each set oftransformations. VR = Vertical Reflection, VS = Vertical Stretch, VT = Vertical Translation, HT = HorizontalTranslation.
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www.ck12.org Chapter 1. Quadratics
TABLE 1.3:
Number V R V S V T HT1. NO 3 −4 −82. YES 2 5 63. YES 1
2 3 −24. NO 1 −2 45. NO 1
4 1 −36. YES 1 −4 07. NO 2 3 18. YES 1
8 0 2
For each of the following graphs, list the transformations of y = x2 and give the equation in vertex form.
9. .
10..
11. .
12. .
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www.ck12.org Chapter 1. Quadratics
For 16-22, graph the following functions and state the domain and range.
16. f (x) = (x−3)2
17. f (x) = x2 +4
18. f (x) = (x+1)2−6
19. f (x) = 3x2
20. f (x) =−34 x2
21. f (x) = 2(x−2)2 +3
22. f (x) = (x−1)2 +2
Calculator Investigation The parent graph of a quadratic equation is y = x2.
23. Graph y = x2,y = 3x2, and y = 12 x2 on the same set of axes in the calculator. Describe how a effects the shape
of the parabola.
24. Graph y = x2,y =−x2, and y =−2x2 on the same set of axes in the calculator. Describe how a effects the shapeof the parabola.
25. Graph y = x2,y = (x− 1)2, and y = (x+ 4)2 on the same set of axes in the calculator. Describe how h effectsthe location of the parabola.
26. Graph y = x2,y = x2 + 2, and y = x2− 5 on the same set of axes in the calculator. Describe how k effects thelocation of the parabola.
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1.3. Vertex Form of a Quadratic Function www.ck12.org
1.3 Vertex Form of a Quadratic Function
Here you will learn to write the equation for a parabola that has undergone transformations.
Given the equation y = 3(x+4)2 +2, list the transformations of y = x2.
Watch This
James Sousa: Find the Equation of a Quadratic Function from a Graph
MEDIAClick image to the left for more content.
Guidance
The equation for a basic parabola with a vertex at (0,0) is y = x2. You can apply transformations to the graph ofy = x2 to create a new graph with a corresponding new equation. This new equation can be written in vertex form.The vertex form of a quadratic function is y = a(x−h)2 + k where:
• |a| is the vertical stretch factor. If a is negative, there is a vertical reflection and the parabola will opendownwards.
• k is the vertical translation.• h is the horizontal translation.
Given the equation of a parabola in vertex form, you should be able to sketch its graph by performing transformationson the basic parabola. This process is shown in the examples.
Example A
Given the following function in vertex form, identify the transformations of y = x2.
y =−12(x−2)2−1
Solution:
• a –Is a negative? YES. The parabola will open downwards.
• a –Is there a number in front of the squared portion of the equation? YES. The vertical stretch factor is theabsolute value of this number. Therefore, the vertical stretch of this function is 1
2 .
• k –Is there a number after the squared portion of the equation? YES. The value of this number is the verticaltranslation. The vertical translation is –1.
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www.ck12.org Chapter 1. Quadratics
• h –Is there a number after the variable ’x’? YES. The value of this number is the opposite of the sign thatappears in the equation. The horizontal translation is +2.
Example B
Given the following transformations, determine the equation of the image of y = x2 in vertex form.
• Vertical stretch by a factor of 3• Vertical translation up 5 units• Horizontal translation left 4 units
Solution:
• a –The image is not reflected in the x-axis. A negative sign is not required.
• a –The vertical stretch is 3, so a = 3.
• k –The vertical translation is 5 units up, so k = 5.
• h –The horizontal translation is 4 units left so h =−4.
The equation of the image of y = x2 is y = 3(x+4)2 +5.
Example C
Using y = x2 as the base function, identify the transformations that have occurred to produce the following imagegraph. Use these transformations to write the equation in vertex form.
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Solution:
a –The parabola does not open downward so a will be positive.
a –The y-values of 1 and 4 are now up 3 and up 12. a = 3.
k –The y−coordinate of the vertex is –5 so k =−5.
h –The x−coordinate of the vertex is +3 so h = 3.
The equation is y = 3(x− 3)2− 5. You can also easily identify the axis of symmetry from the graph which isx = 3. The equation for the axis of symmetry is x =− b
2a .
Example D
In general, the mapping rule used to generate the image of a function is (x,y) → (x′,y′) where (x′,y′) are thecoordinates of the image graph. The resulting mapping rule from y = x2 to the image y = a(x−h)2 + k is (x,y)→(x+h,ay+k). The mapping rule details the transformations that were applied to the coordinates of the base functiony = x2.
Given the following quadratic equation, y = 2(x+3)2 +5 write the mapping rule and create a table of values for themapping rule.
Solution:
The mapping rule for this function will tell exactly what changes were applied to the coordinates of the base quadraticfunction.
y = 2(x+3)2 +5 : (x,y)→ (x−3,2y+5)
These new coordinates of the image graph can be plotted to generate the graph.
Concept Problem Revisited
Given the equation y = 3(x+4)2 +2, list the transformations of y = x2.
a = 3 so the vertical stretch is 3. k = 2 so the vertical translation is up 2. h =−4 so the horizontal translation is left4.
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Vocabulary
Horizontal translationThe horizontal translation is the change in the base graph y = x2 that shifts the graph right or left. It changesthe x−coordinate of the vertex.
Mapping RuleThe mapping rule defines the transformations that have occurred to a function. The mapping rule is (x,y)→(x′,y′) where (x′,y′) are the coordinates of the image graph.
TransformationA transformation is any change in the base graph y = x2. The transformations that apply to the parabola area horizontal translation, a vertical translation, a vertical stretch and a vertical reflection.
Vertex form of y = x2
The vertex form of y = x2 is the form of the quadratic base function y = x2 that shows the transformations ofthe image graph. The vertex form of the equation is y = a(x−h)2 + k.
Vertical ReflectionThe vertical reflection is the reflection of the image graph in the x-axis. The graph opens downward and they-values are negative values.
Vertical StretchThe vertical stretch is the change made to the base function y = x2 by stretching (or compressing) the graphvertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the originalbase graph of y = x2.
Vertical TranslationThe vertical translation is the change in the base graph y = x2 that shifts the graph up or down. It changes they−coordinate of the vertex.
Guided Practice
1. Identify the transformations of y = x2 for the quadratic function −2(y+3) = (x−4)2
2. List the transformations of y = x2 and graph the function =−(x+5)2 +4
3. Graph the function y = 2(x−2)2 +3 using the mapping rule method.
Answers:
1. Rewrite the equation in vertex form. a –a is negative so the parabola opens downwards.
a –The vertical stretch of this function is 12 .
k –The vertical translation is -3.
h –The horizontal translation is +4.
2.
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1.3. Vertex Form of a Quadratic Function www.ck12.org
a→ negative
a→ 1
k→+4
h→−5
3. Mapping Rule (x,y)→ (x+2,2y+3)
Make a table of values:
TABLE 1.4:
x→ x+2 y→ 2y+3–3 –1 9 21–2 0 4 11–1 1 1 50 2 0 31 3 1 52 4 4 113 5 9 21
Draw the graph:
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www.ck12.org Chapter 1. Quadratics
Practice
Identify the transformations of y = x2 in each of the given functions:
1. y = 4(x−2)2−92. y =−1
6 x2 +73. y =−3(x−1)2−64. y = 1
5(x+4)2 +35. y = 5(x+2)2
Graph the following quadratic functions. Give the axis of symmetry and vertex for each function.
6. y = 2(x−4)2−57. y =−1
3(x−2)2 +68. y =−2(x+3)2 +79. y =−1
2(x+6)2 +910. y = 1
3(x−4)2
Using the following mapping rules, write the equation, in vertex form, that represents the image of y = x2.
11. (x,y)→(x+1,−1
2 y)
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1.3. Vertex Form of a Quadratic Function www.ck12.org
12. (x,y)→ (x+6,2y−3)13. (x,y)→
(x−1, 2
3 y+2)
14. (x,y)→ (x+3,3y+1)15. (x,y)→
(x−5,−1
3 y−7)
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www.ck12.org Chapter 1. Quadratics
1.4 Graphs to Solve Quadratic Equations
Here you will learn how to solve a quadratic equation by graphing.
Now that you are familiar with the graph of quadratics, you can use a graph to solve x2−2x−3 = 0.
Watch This
James Sousa: Solve a Quadratic Equation Graphically on the Calculator
MEDIAClick image to the left for more content.
Guidance
Recall that a quadratic equation is a degree 2 equation that can be written in the form ax2 + bx+ c = 0. Everyquadratic equation has a corresponding quadratic function that you get by changing the "0" to a "y". Standardform for a quadratic function is y = ax2 + bx+ c. Quadratic functions can be graphed by hand, or with a graphingcalculator.
How do the solutions to the equation x2 +x−12 = 0 show up on the graph of y = x2 +x−12? On the graph you arelooking for the points that have a y-coordinate that is equal to 0. Therefore, the solutions to the equation will showup as the x-intercepts on the graph of the function. These are also known as the roots or zeros of the function. Hereis the graph of y = x2 + x−12:
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1.4. Graphs to Solve Quadratic Equations www.ck12.org
You can see the x-intercepts are at (−4,0) and (3,0). This means that the solutions to the equation x2 + x−12 = 0are x =−4 and x = 3. You can verify these solutions by substituting them back into the equation:
• (−4)2 +(−4)−12 = 16−4−12 = 0
• (3)2 +(3)−12 = 9+3−12 = 0
Graphing is a great way to solve quadratic equations. Keep in mind that you can also solve many quadratic equationsby factoring or using other algebraic methods such as the quadratic formula or completing the square.
Example A
Solve the following quadratic equation by finding the x-intercepts of the corresponding quadratic function: x2−2x−8 = 0
Solution: The corresponding function is y = x2−2x−8. Use your graphing calculator to make a table and a graphfor this function.
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www.ck12.org Chapter 1. Quadratics
The x-intercepts are (–2, 0) and (4, 0). The x-intercepts are the values for ’x’ that result in y = 0 and are thereforethe solutions to your equation. The solutions for the quadratic are x =−2 and x = 4.
Example B
Solve the following quadratic equation by finding the x-intercepts of the corresponding quadratic function: x2+4x+4 = 0
Solution: The corresponding function is y = x2 +4x+4. Use your graphing calculator to make a table and a graphfor this function.
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1.4. Graphs to Solve Quadratic Equations www.ck12.org
The only x-intercept is (–2, 0). There is only one solution to the equation: x = −2. Keep in mind that quadraticequations can have 0, 1, or 2 real solutions. If you were to factor the quadratic x2+4x+4, you would get (x+2)(x+2)–two of the same factors. The root of −2 for this function is said to have a multiplicity of 2, because 2 factorsproduce the same solution. You will learn more about multiplicity when you study polynomials in future courses.
Example C
Solve the following quadratic equation by finding the x-intercepts of the corresponding quadratic function: x2+3x =10
Solution: First rewrite the equation so it is set equal to zero to get x2+3x−10 = 0. Now, the corresponding functionis y = x2 +3x−10. Use your graphing calculator to make a graph for this function. You will see that there are twox-intercepts.
For this example you will see how the calculator can calculate the zeros of a function on a graph. This technique isparticularly useful when the intercepts are not at whole numbers. Have the calculator find the x-intercept on the leftfirst. Press
The calculator will display “Left Bound?” Use the arrow to position the cursor so that it is to the left and above thex-axis.
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www.ck12.org Chapter 1. Quadratics
When the cursor has been positioned, press
The calculator will now display “Right Bound?” Use the arrows to position the cursor so that it is to the right andbelow the x-axis.
When the cursor has been positioned, press
The calculator will now display “Guess?”
Press
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At the bottom of the screen you can see it says "Zero" and the x and y coordinates. You are interested in the x-coordinate because that is one of the solutions to the original equation. The x-intercept is (–5, 0) which means thatone of the solutions is x =−5.
Repeat this same process to determine the value of the x-intercept on the right. When you do this process for thex-intercept on the right, you will place your cursor below the x-axis for the left bound and above for the right bound.
The x-intercept is (2, 0) which means that the second solution is x = 2.
Concept Problem Revisited
To solve the equation x2 − 2x− 3 = 0 using a graph, use a calculator to graph the corresponding function y =x2−2x−3. Then, look for the values on the graph where y = 0, which will be the x-intercepts.
Another way to think about this problem is to solve the system:
{y = x2−2x−3
y = 0
}
You are looking for where the parabola y = x2−2x−3 intersects with the line y = 0.
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The points of intersection are (–1, 0) and (3, 0). The solutions to the original equation are x =−1 and x = 3.
Vocabulary
Quadratic EquationA quadratic equation is an equation of degree 2. The standard form of a quadratic equation is ax2+bx+c = 0where a 6= 0.
Quadratic FunctionA quadratic function is a function that can be written in the form f (x) = ax2 +bx+ c with a 6= 0. The graphof a quadratic function is a parabola.
Zeros of a Quadratic FunctionThe zeros of a quadratic function are the x-intercepts of the function. These are the values for the variable ’x’that will result in y = 0.
Roots of a Quadratic FunctionThe roots of a quadratic function are also the x-intercepts of the function. These are the values for the variable’x’ that will result in y = 0.
Guided Practice
Solve each quadratic equation using a graph.
1.
x2−3x−10 = 0
2.
2x2−5x+2 = 0
3.
2x2−5x = 3
Answers:
1. To begin, create a table of values for the corresponding function y= x2−3x−10 by using your graphing calculator:
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1.4. Graphs to Solve Quadratic Equations www.ck12.org
From the table, the x-intercepts are (–2, 0) and (5, 0). The x-intercepts are the values for ’x’ that result in y = 0 andare therefore the solutions to the equation.
The solutions to the equation are are x =−2 and x = 5.
2. To begin, create a table of values for the corresponding function y= 2x2−5x+2 by using your graphing calculator:
Press
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www.ck12.org Chapter 1. Quadratics
Press the following keys to determine the x-intercept to the left:
Press the following keys to determine the x-intercept to the right:
The x-intercepts of the function are (0.5, 0) and (2, 0). The solutions to the equation are, therefore, x = 0.5 andx = 2.
3. First rewrite the equation so it is set equal to zero: 2x2− 5x− 3 = 0. Next, create a table of values for thecorresponding function y = 2x2−5x−3 by using your graphing calculator:
Now sketch the graph of the function.
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1.4. Graphs to Solve Quadratic Equations www.ck12.org
The zeros of the function are (–0.5, 0) and (3, 0). Therefore, the solutions to the equation are x =−0.5 and x = 3.
Practice
Use your graphing calculator to solve each of the following quadratic equations by graphing:
1. 2x2 +9x−18 = 02. 3x2 +8x−3 = 03. −5x2 +13x+6 = 04. 2x2−11x+5 = 05. 3x2 +8x−3 = 06. x2− x−20 = 07. 2x2−7x+5 = 08. 3x2 +7x =−29. 2x2−15 =−x
10. 3x2−10x = 811. How could you use the graphs of a system of equations to solve 3x2−10x = 8?12. What’s the difference between a quadratic equation and a quadratic function?13. Will a quadratic equation always have 2 solutions? Explain.14. The quadratic equationx2 +4 = 0 has no real solutions. How does the graph of y = x2 +4 verify this fact?15. When does it make sense to use the graphing method for solving a quadratic equation?16. During practice a Sea World performer starts walking a tightrope at a height of 30 feet but slips and falls into a
safety net 15 feet below. The function h(t) =−16t2 +30 where t represents time measured in seconds, givesthe performer’s height above the ground (in feet) as he falls. Write and solve an equation to find the elapsedtime until the performer lands in the net.
a. Write the equation you need to solve.
b. Write the two equations you will need to use to solve this problem with a graphing calculator. Rewrite theequations in terms of x and y so they are compatible with your graphing calculator.
c. What intervals would be appropriate for your x-and y-axes on your graphing calculator? Explain.
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www.ck12.org Chapter 1. Quadratics
d. Graph the two functions and use your calculator’s trace or intersect feature to find the elapsed time untilthe performer lands in the net. Is your answer exact or approximate?
e. Although the graphs interesect to the left of the y-axis, why is that point irrelevant to this problem?
f. The distance d (in feet), that a falling object travels as a function of time t (in seconds) is given byd(t) = 16t2. Use this fact to explain the model given in the problem, h(t) =−16t2+30. In particular, explainwhy the model includes the constant 30 and why −16t2 includes a negative sign.
g. At what height would the performer have to be for his fall to last exactly one second? Explain.
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1.5 Use Square Roots to Solve QuadraticEquations
Here you’ll learn how to solve quadratic equations in which finding the solutions involves square roots.
What if you had a quadratic equation like 4x2− 9 = 0 in which both terms were perfect squares? How could yousolve such an equation? After completing this Concept, you’ll be able to solve quadratic equations like this one thatinvolve perfect squares.
Watch This
MEDIAClick image to the left for more content.
CK-12 Foundation: 1004S Solving Quadratic Equations Using Square Roots
Guidance
So far you know how to solve quadratic equations by graphing. In this Concept, we’ll examine equations in whichwe can take the square root of both sides of the equation in order to arrive at the result.
Solve Quadratic Equations Involving Perfect Squares
Let’s first examine quadratic equations of the type
x2− c = 0
We can solve this equation by isolating the x2 term: x2 = c
Once the x2 term is isolated we can take the square root of both sides of the equation. Remember that when we takethe square root we get two answers: the positive square root and the negative square root:
x =√
c and x =−√
c
Often this is written as x =±√
c.
Example A
Solve the following quadratic equations:
a) x2−4 = 0
b) x2−25 = 0
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www.ck12.org Chapter 1. Quadratics
Solution
a) x2−4 = 0
Isolate the x2: x2 = 4
Take the square root of both sides: x =√
4 and x =−√
4
The solutions are x = 2 and x =−2.
b) x2−25 = 0
Isolate the x2: x2 = 25
Take the square root of both sides: x =√
25 and x =−√
25
The solutions are x = 5 and x =−5.
We can also find the solution using the square root when the x2 term is multiplied by a constant—in other words,when the equation takes the form
ax2− c = 0
We just have to isolate the x2:
ax2 = b
x2 =ba
Then we can take the square root of both sides of the equation:
x =
√ba
and x =−√
ba
Often this is written as: x =±√
ba
.
Example B
Solve the following quadratic equations.
a) 9x2−16 = 0
b) 81x2−1 = 0
Solution
a) 9x2−16 = 0
Isolate the x2:
9x2 = 16
x2 =169
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1.5. Use Square Roots to Solve Quadratic Equations www.ck12.org
Take the square root of both sides: x =
√169
and x =−√
169
Answer: x = 43 and x =−4
3
b) 81x2−1 = 0
Isolate the x2:
81x2 = 1
x2 =181
Take the square root of both sides: x =
√181
and x =−√
181
Answer: x = 19 and x =−1
9
In some cases, quadratic equations have no real solutions.
Example C
Solve the following quadratic equations.
a) x2 +1 = 0
b) 4x2 +9 = 0
Solution
a) x2 +1 = 0
Isolate the x2: x2 =−1
Take the square root of both sides: x =√−1 and x =−
√−1
Square roots of negative numbers do not give real number results, so there are no real solutions to this equation.
b) 4x2 +9 = 0
Isolate the x2:
4x2 =−9
x2 =−94
Take the square root of both sides: x =
√−9
4and x =−
√−9
4There are no real solutions.
We can also use the square root function in some quadratic equations where both sides of an equation are perfectsquares. This is true if an equation is of this form:
(x−2)2 = 9
Both sides of the equation are perfect squares. We take the square root of both sides and end up with two equations:x−2 = 3 and x−2 =−3.
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www.ck12.org Chapter 1. Quadratics
Solving both equations gives us x = 5 and x =−1.
Example D
Solve the following quadratic equations.
a) (x−1)2 = 4
b) (x+3)2 = 1
Solution
a) (x−1)2 = 4
Take the square root of both sides : x−1 = 2 and x−1 =−2
Solve each equation : x = 3 and x =−1
Answer: x = 3 and x =−1
b) (x+3)2 = 1
Take the square root of both sides : x+3 = 1 and x+3 =−1
Solve each equation : x =−2 and x =−4
Answer: x =−2 and x =−4
It might be necessary to factor the right-hand side of the equation as a perfect square before applying the methodoutlined above.
Watch this video for help with the Examples above.
MEDIAClick image to the left for more content.
CK-12 Foundation: 1004 Solving Quadratic Equations Using Square Roots
Vocabulary
• The solutions of a quadratic equation are often called the roots or zeros.
Practice
Solve the following quadratic equations. If there is no real solution, write no real solution.
1. x2−1 = 02. x2−100 = 03. x2 +16 = 0
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1.5. Use Square Roots to Solve Quadratic Equations www.ck12.org
4. 9x2−1 = 05. 4x2−49 = 06. 64x2−9 = 07. x2−81 = 08. 25x2−36 = 09. x2 +9 = 0
10. x2−16 = 011. x2−36 = 012. 16x2−49 = 013. (x−2)2 = 114. (x+5)2 = 1615. (2x−1)2−4 = 016. −5(x+3)2 =−8017. 2(x+3)2 = 818. 0.5(x+2)2−4 = 1419. 3(x−1)2 +1 = 1920. To study how high a ball bounces, students drop the ball from various heights. The function h(t) =−16t2+h0
gives the height (in feet) of the ball at time t measured in seconds since the ball was dropped from a height ofh0.
a. The ball is dropped from a height of h0 = 12 f t. Write and solve an equation to find the elapsed time untilthe ball hits the floor (round your answer to the nearest hundredth, as necessary).
b. Does doubling the drop height also double the elapsed time until the ball hits the floor? Explain yourreasoning.
c. When dropped from a height of h0 = 24 f t, the ball rebounds to a height of 12 feet and then falls back tothe floor. Find the total time for this to happen. (Assume the ball takes the same time to rebound 12 feet asit does to fall 12 feet; round your answer to the nearest hundredth, as necessary).
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1.6 Solving Quadratics using Factoring
Here you’ll learn how to solve factorable quadratic equations for x.
The height of a ball that is thrown straight up in the air from a height of 2 meters above the ground with a velocity of9 meters per second is given by the quadratic equation h =−5t2 +9t +2, where t is the time in seconds. How longdoes it take the ball to hit the ground?
Watch This
Watch the first part of this video, until about 4:40.
MEDIAClick image to the left for more content.
Khan Academy: Solving Quadratic Equations by Factoring.avi
Guidance
In the previous unit, you worked on factoring quadratics. Now, we will apply factoring to solving a quadraticequation. It adds one additional step to the end of what you did in the previous unit. Let’s go through an example.
Example A
Solve x2−9x+18 = 0 by factoring.
Solution: The only difference between this problem and previous ones from the concepts before is the addition ofthe = sign. Now that this is present, we need to solve for x. We can still factor the way we always have. Becausea = 1, determine the two factors of 18 that add up to -9.
x2−9x+18 = 0
(x−6)(x−3) = 0
Now, we have two factors that, when multiplied, equal zero. Recall that when two numbers are multiplied togetherand one of them is zero, the product is always zero.
Zero-Product Property: If ab = 0, then a = 0 or b = 0.
This means that x− 6 = 0 OR x− 3 = 0. Therefore, x = 6 or x = 3. There will always be the same number ofsolutions as factors.
Check your answer:
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1.6. Solving Quadratics using Factoring www.ck12.org
62−9(6)+18 = 0 or 32−9(3)+18 = 0
36−54+18 = 0 9−27+18 = 0
Example B
Solve 6x2 + x−4 = 11 by factoring.
Solution: At first glance, this might not look factorable to you. However, before we factor, we must combine liketerms. Also, the Zero-Product Property tells us that in order to solve for the factors, one side of the equation must bezero.
6x2 + x−4 =ZZ11
−11 =−ZZ11
6x2 + x−15 = 0
Now, factor. The product of ac is -90. What are the two factors of -90 that add up to 1? 10 and -9. Expand thex−term and factor.
6x2 + x−15 = 0
6x2−9x+10x−15 = 0
3x(2x−3)+5(2x−3) = 0
(2x−3)(3x+5) = 0
Lastly, set each factor equal to zero and solve.
2x−3 = 0 3x+5 = 0
2x = 3 or 3x =−5
x =32
x =−53
Check your work:
6(
32
)2
+32−4 = 11 6
(−5
3
)2
− 53−4 = 11
6 · 94+
32−4 = 11 or 6 · 25
9− 5
3−4 = 11
272+
32−4 = 11
503− 5
3−4 = 11
15−4 = 11 15−4 = 11
Example C
Solve 10x2−25x = 0 by factoring.
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Solution: Here is an example of a quadratic equation without a constant term. The only thing we can do is take outthe GCF.
10x2−25x = 0
5x(2x−5) = 0
Set the two factors equal to zero and solve.
5x = 0 2x−5 = 0
x = 0 or 2x = 5
x =52
Check:
10(0)2−25(0) = 0 10(
52
)2
−25(
52
)= 0
0 = 0 or 10 · 254− 125
2= 0
1252− 125
2= 0
Intro Problem Revisit When the ball hits the ground, the height h is 0. So the equation becomes 0 =−5t2 +9t +2.
Let’s factor and solve for t. −5t2 +9t +2
We need to find the factors of −10 that add up to 9. Testing the possibilities, we find 10 and -1 to be the correctcombination.
−5t2 +10t− t +2 = (−5t2 +10t)+(−t +2) = 5t(−t +2)+(−t +2) = (5t +1)(−t +2)
Now set this factorization equal to zero and solve.
(5t +1)(−t +2) = 0
Because t represents the time, it must be positive. Only (−t +2) = 0 results in a positive value.
t = 2, therefore it takes the ball 2 seconds to reach the ground.
Guided Practice
Solve the following equations by factoring.
1. 4x2−12x+9 = 0
2. x2−5x = 6
3. 8x−20x2 = 0
4. 12x2 +13x+7 = 12−4x
Answers
1. ac = 36. The factors of 36 that also add up to -12 are -6 and -6. Expand the x−term and factor.
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4x2−12x+9 = 0
4x2−6x−6x+9 = 0
2x(2x−3)−3(2x−3) = 0
(2x−3)(2x−3) = 0
The factors are the same. When factoring a perfect square trinomial, the factors will always be the same. In thisinstance, the solutions for x will also be the same. Solve for x.
2x−3 = 0
2x = 3
x =32
When the two factors are the same, we call the solution for x a double root because it is the solution twice.
2. Here, we need to get everything on the same side of the equals sign in order to factor.
x2−5x = 6
x2−5x−6 = 0
Because there is no number in front of x2, we need to find the factors of -6 that add up to -5.
(x−6)(x+1) = 0
Solving each factor for x, we get that x = 6 or x =−1.
3. Here there is no constant term. Find the GCF to factor.
8x−20x2 = 0
4x(2−5x) = 0
Solve each factor for x.
4x = 0 2−5x = 0
x = 0 or 2 = 5x25= x
4. This problem is slightly more complicated than #2. Combine all like terms onto the same side of the equals signso that one side is zero.
12x2 +13x+7 = 12−4x
12x2 +17x−5 = 0
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www.ck12.org Chapter 1. Quadratics
ac =−60. The factors of -60 that add up to 17 are 20 and -3. Expand the x−term and factor.
12x2 +17x−5 = 0
12x2 +20x−3x−5 = 0
4x(3x+5)−1(3x+5) = 0
(3x+5)(4x−1) = 0
Solve each factor for x.
3x+5 = 0 4x−1 = 0
3x =−5 or 4x = 1
x =−53
x =14
Vocabulary
SolutionThe answer to an equation. With quadratic equations, solutions can also be called zeros or roots.
Double RootA solution that is repeated twice.
Practice
Solve the following quadratic equations by factoring, if possible.
1. x2 +8x−9 = 02. x2 +6x = 03. 2x2−5x = 124. 12x2 +7x−10 = 05. x2 = 96. 30x+25 =−9x2
7. 2x2 + x−5 = 08. 16x = 32x2
9. 3x2 +28x =−3210. 36x2−48 = 111. 6x2 + x = 412. 5x2 +12x+4 = 0
Challenge Solve these quadratic equations by factoring. They are all factorable.
13. 8x2 +8x−5 = 10−6x14. −18x2 = 48x+1415. 36x2−24 = 96x−3916. Real Life Application George is helping his dad build a fence for the backyard. The total area of their
backyard is 1600 square feet. The width of the house is half the length of the yard, plus 7 feet. How muchfencing does George’s dad need to buy?
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17. The width of a high school soccer field is 45 yards shorter than its length. The area of the field is 9000 squareyards. Find the dimensions.
18. The length of a rectangular swimming pool is 20 feet greater than its width. The area of the pool is 525 squarefeet. What are the length and width of the pool? Are there any solutions to the equation you solved that don’tmake sense in the context of the problem? Explain.
19. An athlete throws a shot put with an initial velocity of 29 feet per second and from an initial height of 6feet. After how many seconds will the shot put hit the ground? (click here for more information aboutvertical motion)
20. When Bobby shoots a free throw, the ball is 6 feet from the floor and has an initial velocity of 20 feet persecond. The hoop is 10 feet from the floor.
a. Use the vertical motion model h(t) =−16t2 + v0t +h0 (click here for more information) to determine anequation that models Bobby’s free throw.
b. How long is the basketball in the air before it reaches the hoop?
c. Raymond shoots a free throw that is 5 foot 9 inches from the floor with the same initial velocity. Will theball be in the air more or less time? Explain.
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www.ck12.org Chapter 1. Quadratics
1.7 Completing the Square
Here you’ll learn how to complete the square to help you solve quadratic equations. You’ll also solve quadraticequations in standard form.
What if you had a quadratic equation like x2 +12x = 13? How could you solve it by taking the square root of bothsides? After completing this Concept, you’ll be able to complete the square to solve quadratic equations like thisone.
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CK-12 Foundation: 1006S Solving Quadratic Equations by Completing the Square
Guidance
You saw in a previous section that if you have a quadratic equation of the form (x−2)2 = 5, you can easily solve itby taking the square root of each side:
x−2 =√
5 and x−2 =−√
5
Simplify to get:
x = 2+√
5≈ 4.24 and x = 2−√
5≈−0.24
So what do you do with an equation that isn’t written in this nice form? In this section, you’ll learn how to rewriteany quadratic equation in this form by completing the square.
Complete the Square of a Quadratic Expression
Completing the square lets you rewrite a quadratic expression so that it contains a perfect square trinomial that youcan factor as the square of a binomial.
Remember that the square of a binomial takes one of the following forms:
(x+a)2 = x2 +2ax+a2
(x−a)2 = x2−2ax+a2
So in order to have a perfect square trinomial, we need two terms that are perfect squares and one term that is twicethe product of the square roots of the other terms.
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1.7. Completing the Square www.ck12.org
Example A
Complete the square for the quadratic expression x2 +4x.
Solution
To complete the square we need a constant term that turns the expression into a perfect square trinomial. Since themiddle term in a perfect square trinomial is always 2 times the product of the square roots of the other two terms,we re-write our expression as:
x2 +2(2)(x)
We see that the constant we are seeking must be 22 :
x2 +2(2)(x)+22
Answer: By adding 4 to both sides, this can be factored as: (x+2)2
Notice, though, that we just changed the value of the whole expression by adding 4 to it. If it had been an equation,we would have needed to add 4 to the other side as well to make up for this.
Also, this was a relatively easy example because a, the coefficient of the x2 term, was 1. When that coefficientdoesn’t equal 1, we have to factor it out from the whole expression before completing the square.
Example B
Complete the square for the quadratic expression 4x2 +32x.
Solution
Factor the coefficient of the x2 term:
4(x2 +8x)
Re-write the expression:
4(x2 +2(4)(x))
We complete the square by adding the constant 42:
4(x2 +2(4)(x)+42)
Factor the perfect square trinomial inside the parenthesis:
4(x+4)2
The expression “completing the square” comes from a geometric interpretation of this situation. Let’s revisit thequadratic expression in Example 1: x2 +4x.
We can think of this expression as the sum of three areas. The first term represents the area of a square of side x.The second expression represents the areas of two rectangles with a length of 2 and a width of x:
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www.ck12.org Chapter 1. Quadratics
We can combine these shapes as follows:
We obtain a square that is not quite complete. To complete the square, we need to add a smaller square of side length2.
We end up with a square of side length (x+ 2); its area is therefore (x+ 2)2. Let’s demonstrate the method ofcompleting the square with an example.
Example C
Solve the following quadratic equation: 3x2−10x =−1
Solution
Divide all terms by the coefficient of the x2 term:
x2− 103
x =−13
Rewrite: x2−2(5
3
)(x) =−1
3
In order to have a perfect square trinomial on the right-hand-side we need to add the constant(5
3
)2. Add this constant
to both sides of the equation:
x2−2(
53
)(x)+
(53
)2
=−13+
(53
)2
Factor the perfect square trinomial and simplify:
(x− 5
3
)2
=−13+
259(
x− 53
)2
=229
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1.7. Completing the Square www.ck12.org
Take the square root of both sides:
x− 53=
√229
and x− 53=−
√229
x =53+
√229≈ 3.23 and x =
53−√
229≈ 0.1
Answer: x = 3.23 and x = 0.1
Solving Quadratic Equations in Standard Form
If an equation is in standard form (ax2 +bx+ c = 0), we can still solve it by the method of completing the square.All we have to do is start by moving the constant term to the right-hand-side of the equation.
Example D
Solve the following quadratic equation: x2 +15x+12 = 0
Solution
Move the constant to the other side of the equation:
x2 +15x =−12
Rewrite: x2 +2(15
2
)(x) =−12
Add the constant(15
2
)2to both sides of the equation:
x2 +2(
152
)(x)+
(152
)2
=−12+(
152
)2
Factor the perfect square trinomial and simplify:
(x+
152
)2
=−12+225
4(x+
152
)2
=1774
Take the square root of both sides:
x+152
=
√1774
and x+152
=−√
1774
x =−152+
√1774≈−0.85 and x =−15
2−√
1774≈−14.15
Answer: x =−0.85 and x =−14.15
Watch this video for help with the Examples above.
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CK-12 Foundation: 1006 Solving Quadratic Equations by Completing the Square
Vocabulary
• A perfect square trinomial has the form a2 +2(ab)+b2, which factors into (a+b)2.
Guided Practice
Solve the following quadratic equation: −x2 +22x = 5
Solution
Divide all terms by the coefficient of the x2 term:
x2−22x =−6
Rewrite: x2−2(11)(x) =−6.
In order to have a perfect square trinomial on the right-hand-side we need to add the constant (11)2. Add thisconstant to both sides of the equation:
x2−2(11)(x)+(11)2 =−6+(11)2
Factor the perfect square trinomial and simplify:
(x−11)2 =−6+(11)2(x− 5
3
)2
= 16
Take the square root of both sides:
x−11 =√
16 and x−11 =−√
16
x = 11+√
16 = 15 and x = 11−√
4 = 7
Answer: x = 15 and x = 7
Practice
Complete the square for each expression.
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1.7. Completing the Square www.ck12.org
1. x2 +5x2. x2−2x3. x2 +3x4. x2−4x5. 3x2 +18x6. 2x2−22x7. 8x2−10x8. 5x2 +12x
Solve each quadratic equation by completing the square.
9. x2−4x = 510. x2−5x = 1011. x2 +10x+15 = 012. x2 +15x+20 = 013. 2x2−18x = 314. 4x2 +5x =−115. 10x2−30x−8 = 016. 5x2 +15x−40 = 0
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1.8 Solving Quadratic Equations by theQuadratic Formula
Learning objectives
• Solve quadratic equations using the quadratic formula.• Identify and choose methods for solving quadratic equations.• Solve real-world problems using functions by completing the square.
Introduction
In this section, you will solve quadratic equations using the Quadratic Formula. It is probably the most usedmethod for solving quadratic equations. For a quadratic equation in standard form
ax2 +bx+ c = 0
The solutions are found using the following formula.
x =−b±
√b2−4ac
2a
We will start by explaining where this formula comes from and then show how it is applied. This formula is derivedby solving a general quadratic equation using the method of completing the square that you learned in the previoussection.
Divide by the coefficient of the x2 term: x2 +ba
x =− ca
Rewrite: x2 +2(
b2a
)x =− c
a
Add the constant(
b2a
)2
to both sides: x2 +2(
b2a
)x+(
b2a
)2
=− ca+
b2
4a2
Factor the perfect square trinomial:(
x+b2a
)2
=−4ac4a2 +
b2
4a2
Simplify:(
x+b2a
)2
=b2−4ac
4a2
Take the square root of both sides: x+b
2a=
√b2−4ac
4a2 and x+b2a
=−√
b2−4ac4a2
Simplify: x+b
2a=
√b2−4ac
2aand x+
b2a
=−√
b2−4ac2a
x =− b2a
+
√b2−4ac
2aand x =− b
2a−√
b2−4ac2a
x =−b+
√b2−4ac
2aand x =
−b−√
b2−4ac2a
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1.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
This can be written more compactly as x = −b±√
b2−4ac2a .
You can see that the familiar formula comes directly from applying the method of completing the square. Applyingthe method of completing the square to solve quadratic equations can be tedious. The quadratic formula is a morestraightforward way of finding the solutions.
Solve Quadratic Equations Using the Quadratic Formula
Applying the quadratic formula basically amounts to plugging the values of a,b and c into the quadratic formula.
Example 1
Solve the following quadratic equation using the quadratic formula.
a) 2x2 +3x+1 = 0
b) x2−6x+5 = 0
c) −4x2 + x+1 = 0
Solution
Start with the quadratic formula and plug in the values of a,b and c.
a)
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a = 2,b = 3,c = 1. x =−3±
√(3)2−4(2)(1)2(2)
Simplify. x =−3±
√9−8
4=−3±
√1
4
Separate the two options. x =−3+1
4and x =
−3−14
Solve. x =−24
=−12
and x =−44
=−1
Answer x =−12 and x =−1
Remember you can check this solution by determine the x-intercepts of the quadratic function y = 2x2 +3x+1
b)
Quadratic formula. x =−b±
√b2−4ac
2a
Plug in the values a = 1,b =−6,c = 5. x =−(−6)±
√(−6)2−4(1)(5)2(1)
Simplify. x =6±√
36−202
=6±√
162
Separate the two options. x =6+4
2and x =
6−42
Solve x =102
= 5 and x =22= 1
Answer x = 5 and x = 1
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www.ck12.org Chapter 1. Quadratics
c)
Quadratic formula. x =−b±
√b2−4ac
2a
Plug in the values a =−4,b = 1,c = 1. x =−1±
√(1)2−4(−4)(1)2(−4)
Simplify. x =−1±
√1+16
−8=−1±
√17
−8
Separate the two options. x =−1+
√17
−8and x =
−1−√
17−8
Solve. x≈−.39 and x≈ .64
Answer x = −1±√
17−8 (This is the exact answer. Decimal approximations are useful in real world applications).
Often when we plug the values of the coefficients into the quadratic formula, we obtain a negative number inside thesquare root. Since the square root of a negative number does not give real answers, we say that the equation has noreal solutions. In more advanced mathematics classes, you will learn how to work with “complex” (or “imaginary”)solutions to quadratic equations.
Example 2
Solve the following quadratic equation using the quadratic formula x2 +2x+7 = 0
Solution:
a)
Quadratic formula. x =b±√
b2−4ac2a
Plug in the values a = 1,b = 2,c = 7. x =−2±
√(2)2−4(1)(7)2(1)
Simplify. x =−2±
√4−28
2=−2±
√−24
2
Answer There are no real solutions.
To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems,we must rewrite the equation before we apply the quadratic formula.
Example 3
Solve the following quadratic equation using the quadratic formula.
a) x2−6x = 10
b) 8x2 +5x =−6
Solution:
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1.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
a)
Rewrite the equation in standard form. x2−6x−10 = 0
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a = 1,b =−6,c =−10. x =−(−6)±
√(−6)2−4(1)(−10)
2(1)
Simplify. x =6±√
36+402
=6±√
762
Separate the two options. x =6+√
762
and x =6−√
762
Solve. x≈ 7.36 and x≈−1.36
Answer: We should give the exact solution. x = 6±√
762
b)
Rewrite the equation in standard form. 8x2 +5x+6 = 0
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a = 8,b = 5,c = 6. x =−5±
√(5)2−4(8)(6)2(8)
Simplify. x =−5±
√25−192
16=−5±
√−167
16
Answer No real solutions.
Notice if we try to check this solution by graphing the quadratic function y = 8x2 +5x+6, the graph does not crossthe x-axis or have x-intercepts. This verifies we have complex solutions with an imaginary part.
Finding the Vertex of a Parabola with the Quadratic Formula
Sometimes you get more information from a formula beyond what you were originally seeking. In this case, thequadratic formula also gives us an easy way to locate the vertex of a parabola.
First, recall that the quadratic formula tells us the roots or solutions of the equation ax2 +bx+ c = 0. Those rootsare
x =−b±
√b2−4ac
2a.
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www.ck12.org Chapter 1. Quadratics
We can rewrite the fraction in the quadratic formula as
x =− b2a±√
b2−4ac2a
.
Recall that the roots are symmetric about the vertex. In the form above, we can see that the roots of a quadratic
equation are symmetric around the x−coordinate− b2a because they move
√b2−4ac
2a units to the left and right (recallthe ± sign) from the vertical line x = − b
2a . The image to the right illustrates this for the equation x2− 2x− 3 = 0.The roots, -1 and 3 are both 2 units from the vertical line x = 1.
Identify and Choose Methods for Solving Quadratic Equations.
In mathematics, you will need to solve quadratic equations that describe application problems or that are part ofmore complicated problems. You learned four ways of solving a quadratic equation.
• Factoring.• Taking the square root.• Completing the square.• Quadratic formula.
Usually you will not be told which method to use. You will have to make that decision yourself. However, here aresome guidelines to which methods are better in different situations.
Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you canfactor because this is the fastest method.
Taking the square root is best used when there is no x term in the equation.
Completing the square can be used to solve any quadratic equation. This is usually not any better than usingthe quadratic formula (in terms of difficult computations), however it is a very important method for re-writing aquadratic function in vertex form. It is also used to re-write the equations of circles, ellipses and hyperbolas instandard form (something you will do in algebra II, trigonometry, physics, calculus, and beyond...).
Quadratic formula is the method that is used most often for solving a quadratic equation if solving directly bytaking square root and factoring does not work.
If you are using factoring or the quadratic formula make sure that the equation is in standard form.
Example 4
Solve each quadratic equation
a) x2−4x−5 = 0
b) x2 = 8
c) −4x2 + x = 2
d) 25x2−9 = 0
e) 3x2 = 8x
Solution
a) This expression if easily factorable so we can factor and apply the zero-product property:
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1.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
Factor. (x−5)(x+1) = 0
Apply zero-product property. x−5 = 0 and x+1 = 0
Solve. x = 5 and x =−1
Answer x = 5 and x =−1
Note that the factored form can also help us graph the function. The values above are also the x-intercepts: (5,0)and (−1,0). After graphing the x-intercepts we can find the equation of the axis of symmetry as well as the vertex.
b) Since the expression is missing the x term we can take the square root:
Take the square root of both sides. x =√
8 and x =−√
8
Answer x =±√
8
c) Rewrite the equation in standard form.
It is not apparent right away if the expression is factorable, so we will use the quadratic formula.
Quadratic formula x =−b±
√b2−4ac
2a
Plug in the values a =−4,b = 1,c =−2. x =−1±
√12−4(−4)(−2)2(−4)
Simplify. x =−1±
√1−32
−8=−1±
√−31
−8
Answer No real solutions.
d) This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in thiscase.
Add 9 to both sides of the equation. 25x2 = 9
Divide both sides by 25. x2 =9
25
Take the square root of both sides. x =
√925
and x =−√
925
Simplify. x =35
and x =−35
Answer x = 35 and x =−3
5
e)
Rewrite the equation in standard form 3x2−8x = 0
Factor out common x term. x(3x−8) = 0
Set both terms to zero. x = 0 and 3x = 8
Solve. x = 0 and x =83
Answer x = 0 and x = 83
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Seeing structure in equations
The form of the quadratic equation can also highlight different variables of interest. Consider the profit, P, (inthousands of dollars), that a company makes selling an item is a quadratic function of the price, x (in dollars), thatthey charge for an item. The following expressions of P(x) are equivalent:
Form 1: P(x) =−2x2 +12x−16
Form 2: P(x) =−2(x−2)(x−4)
Form 3: P(x) =−2(x−3)2 +2
Which of the equivalent expressions for P(x) reveals the price which gives a profit of zero without changing theform of the expression?
The factored form, Form 2, reveals the price which gives a profit of zero, which can clearly be seen with thefollowing:
0 =−2(x−2)(x−4)
Since x = 2 or x = 4, the company breaks even if the price charged for the product is $2 or $4.
Which of the equivalent expressions for P(x) reveals the profit when the price is zero without changing the formof the expression?
The quadratic function in standard form, or Form 1, would reveal the profit when the price is zero. This can be seenby substituting 0 in for x:
P(0) =−2(0)2 +12(0)−16
In this case, c =−16 so the profit is −16 (in thousands of dollars) when the price is zero. If the company gives theproduct away for free, it loses $16,000.
Which of the equivalent expressions for P(x) reveals the price which produces the highest possible profit withoutchanging the form of the expression?
Vertex form, or Form 3, reveals the maximum profit without any computation. From P(x) =−2(x−3)2 +2 we cansee the maximum profit is 2 thousand dollars and it occurs when x = 3. The company should charge a price of $3for this product in order to maximize its profits.
Solve Real-World Problems Using Quadratic Functions by any Method
Here are some application problems that arise from number relationships and geometry applications.
Example 5
The product of two positive consecutive integers is 156. Find the integers.
Solution
For two consecutive integers, one integer is one more than the other one.
Define
Let x = the smaller integer
x+1 = the next integer
Translate
The product of the two numbers is 156. We can write the equation:
x(x+1) = 156
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1.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
Solve
x2 + x = 156
x2 + x−156 = 0
Apply the quadratic formula with a = 1,b = 1,c =−156
x =−1±
√12−4(1)(−156)
2(1)
x =−1±
√625
2=−1±25
2
x =−1+25
2and x =
−1−252
x =242
= 12 and x =−26
2=−13
Since we are looking for positive integers take, x = 12
Answer 12 and 13
Check 12×13 = 156. The answer checks out.
Example 6
The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square/meters. Findthe dimensions of the pool.
Solution:
Draw a sketch
Define
Let x = the width of the pool
x+10 = the length of the pool
Translate
The area of a rectangle is A = length × width, so
x(x+10) = 875
Solve
x2 +10x = 875
x2 +10x−875 = 0
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www.ck12.org Chapter 1. Quadratics
Apply the quadratic formula with a = 1,b = 10 and c =−875
x =−10±
√(10)2−4(1)(−875)
2(1)
x =−10±
√100+35002
x =−10±
√3600
2=−10±60
2
x =−10+60
2and x =
−10−602
x =502
= 25 and x =−70
2=−35
Since the dimensions of the pools should be positive, then x = 25 meters.
Answer The pool is 25 meters×35 meters.
Check 25×35 = 875 m2. The answer checks out.
Example 7
Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the wholegarden and between each section as shown. The plot is twice as long as it is wide and the total area is 200 f t2. Howmuch fencing does Suzie need?
Solution
Draw a Sketch
Define
Let x = the width of the plot
2x = the length of the plot
Translate
Area of a rectangle is A = length × width, so
x(2x) = 200
Solve
2x2 = 200
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1.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
Solve by taking the square root.
x2 = 100
x =√
100 and x =−√
100
x = 10 and x =−10
We take x = 10 since only positive dimensions make sense.
The plot of land is 10 f eet×20 f eet.
To fence the garden the way Suzie wants, we need 2 lengths and 4 widths = 2(20)+4(10) = 80 f eet of fence.
Answer: The fence is 80 feet.
Check 10×20 = 200 f t2 and 2(20)+4(10) = 80 f eet. The answer checks out.
Example 8
An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and thetip of the triangle touches the opposite side of the square. If the area of the triangle is 20 in2 what is the area of thesquare?
Solution:
Draw a sketch.
Define
Let x = base of the triangle
x = height of the triangle
Translate
Area of a triangle is 12 ×base×height, so
12· x · x = 20
Solve
12
x2 = 20
Solve by taking the square root.
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www.ck12.org Chapter 1. Quadratics
x2 = 40
x =√
40 = 2√
10 and x =−√
40 =−2√
10
x≈ 6.32 and x≈−6.32
The side of the square is approximately 6.32 inches.
The area of the square is (6.32)2 ≈ 40 in2, twice as big as the area of the triangle.
Answer: Area of the triangle is 40 in2
Check: It makes sense that the area of the square will be twice that of the triangle. If you look at the figure you cansee that you can fit two triangles inside the square.
FIGURE 1.1
The answer checks.
Review Questions
Solve the following quadratic equations using the quadratic formula.
1. x2 +4x−21 = 02. x2−6x = 123. 3x2− 1
2 x = 38
4. 2x2 + x−3 = 05. −x2−7x+12 = 06. −3x2 +5x = 07. 4x2 = 08. x2 +2x+6 = 0
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1.8. Solving Quadratic Equations by the Quadratic Formula www.ck12.org
Solve the following quadratic equations using the method of your choice.
9. x2− x = 6
10. x2−12 = 0
11. −2x2 +5x−3 = 0
12. x2 +7x−18 = 0
13. 3x2 +6x =−10
14. −4x2 +4000x = 0
15. −3x2 +12x+1 = 0
16. x2 +6x+9 = 0
17. 81x2 +1 = 0
18. −4x2 +4x = 9
19. 36x2−21 = 0
20. x2−2x−3 = 0
21. The product of two consecutive integers is 72. Find the two numbers.
22. The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers.
23. The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, find itsdimensions.
24. Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece ofwood is 4 f eet×8 f eet and the cut off part is 1
3 of the total area of the plywood sheet. What is the length of the sideof the square?
25. Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of thepatio is 192 f t2 and the length is 4 feet longer than the width. Find how much fencing Mike will need.
26. The Profit, P (in thousands of dollars), that a company makes selling an item is a quadratic function of theprice, x (in dollars), that they charge for the item. The following quadratic function is used to show the price of theproduct which gives a profit of zero: P(x) =−4(x−4)(x−6).
a. Find the price(s) that give a profit of zero.
b. Create an equivalent quadratic expression that clearly shows the profit when the price is zero. What is theprofit when the price is zero?
c. Create an equivalent quadratic expression that clearly shows the price which produces the highest possibleprofit. What is the price and what is the highest possible profit at this price?
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1.9 Solutions Using the Discriminant
Here you’ll learn how to find the number of solutions to a quadratic equation by calculating the discriminant andhow to use this information to solve problems.
Suppose that the balance of your checking account in dollars can be modeled by the function B(t) = 0.001t2− t +300, where t is the number of days the checking account has been open. Will the balance of your checking accountever be $100? In this Concept, you’ll learn how to find the discriminant of a quadratic equation so that you can findthe number of solutions to the equation and answer real-world questions such as this one.
Guidance
You have seen parabolas that intersect the x−axis twice, once, or not at all. There is a relationship between thenumber of real x−intercepts and the quadratic formula.
Case 1: The parabola has two x−intercepts. This situation has two possible solutions for x, because the value inside
the square root is positive. Using the quadratic formula, the solutions are x= −b+√
b2−4ac2a and x= −b−
√b2−4ac2a .
Case 2: The parabola has one x−intercept. This situation occurs when the vertex of the parabola just touches thex−axis. This is called a repeated root, or double root. The value inside the square root is zero. Using the quadraticformula, the solution is x = −b
2a .
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Case 3: The parabola has no x−intercept. This situation occurs when the parabola does not cross the x−axis.The value inside the square root is negative, so there are no real roots. The solutions to this type of situation areimaginary, which you will learn more about in a later textbook.
The value inside the square root of the quadratic formula is called the discriminant. It is symbolized by D. Itdictates the number of real solutions the quadratic equation has. This can be summarized with the DiscriminantTheorem.
• If D > 0, the parabola will have two x−intercepts. The quadratic equation will have two real solutions.• If D = 0, the parabola will have one x−intercept. The quadratic equation will have one real solution.• If D < 0, the parabola will have no x−intercepts. The quadratic equation will have zero real solutions.
Example A
Determine the number of real solutions to −3x2 +4x+1 = 0.
Solution: By finding the value of its discriminant, you can determine the number of x−intercepts the parabola hasand thus the number of real solutions.
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D = b2−4(a)(c)
D = (4)2−4(−3)(1)
D = 16+12 = 28
Because the discriminant is positive, the parabola has two real x−intercepts and thus two real solutions.
Example B
Determine the number of solutions to −2x2 + x = 4.
Solution: Before we can find its discriminant, we must write the equation in standard form: ax2 +bx+ c = 0.
Subtract 4 from each side of the equation: −2x2 + x−4 = 0.
Find the discriminant: D = (1)2−4(−2)(−4)
D = 1−32 =−31
The value of the discriminant is negative; there are no real solutions to this quadratic equation. The parabola doesnot cross the x−axis.
Example C
Emma and Bradon own a factory that produces bike helmets. Their accountant says that their profit per year is givenby the function P = 0.003x2 + 12x+ 27,760, where x represents the number of helmets produced. Their goal is tomake a profit of $40,000 this year. Is this possible?
Solution: The equation we are using is 40,000 = 0.003x2 +12x+27,760. By finding the value of its discriminant,you can determine if the profit is possible.
Begin by writing this equation in standard form:
0 = 0.003x2 +12x−12,240
D = b2−4(a)(c)
D = (12)2−4(0.003)(−12,240)
D = 144+146.88 = 290.88
Because the discriminant is positive, the parabola has two real solutions. Yes, the profit of $40,000 is possible.
Video Review
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Guided Practice
Determine the number of solutions for x2−2x+1 = 0.
Solution:
Substitute the values into the discriminant:
D = b2−4(a)(c)
D = (−2)2−4(1)(1)
D = 4−4 = 0
Because the discriminant is zero, the parabola has one real x−intercept and thus one real solution.
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Notethat there is not always a match between the number of the practice exercise in the video and the number of thepractice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Discriminant of Quadratic Equations (10:14)
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1. What is a discriminant? What does it do?2. What is the formula for the discriminant?3. Can you find the discriminant of a linear equation? Explain your reasoning.4. Suppose D = 0. Draw a sketch of this graph and determine the number of real solutions.5. D = −2.85. Draw a possible sketch of this parabola. What is the number of real solutions to this quadratic
equation.6. D > 0. Draw a sketch of this parabola and determine the number of real solutions.
Find the discriminant of each quadratic equation.
7. 2x2−4x+5 = 08. x2−5x = 8
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9. 4x2−12x+9 = 010. x2 +3x+2 = 011. x2−16x = 3212. −5x2 +5x−6 = 0
Determine the nature of the solutions of each quadratic equation.
13. −x2 +3x−6 = 014. 5x2 = 6x15. 41x2−31x−52 = 016. x2−8x+16 = 017. −x2 +3x−10 = 018. x2−64 = 0
A solution to a quadratic equation will be irrational if the discriminant is not a perfect square. If the discriminant isa perfect square, then the solutions will be rational numbers. Using the discriminant, determine whether the solutionswill be rational or irrational.
19. x2 =−4x+2020. x2 +2x−3 = 021. 3x2−11x = 1022. 1
2 x2 +2x+ 23 = 0
23. x2−10x+25 = 024. x2 = 5x25. Marty is outside his apartment building. He needs to give Yolanda her cell phone but he does not have time
to run upstairs to the third floor to give it to her. He throws it straight up with a vertical velocity of 55feet/second. Will the phone reach her if she is 36 feet up? (Hint: The equation for the height is given byy =−32t2 +55t +4.)
26. Bryson owns a business that manufactures and sells tires. The revenue from selling the tires in the month ofJuly is given by the function R = x(200− 0.4x) where x is the number of tires sold. Can Bryson’s businessgenerate revenue of $20,000 in the month of July?
27. Marcus kicks a football in order to score a field goal. The height of the ball is given by the equation y =− 32
6400 x2 +x, where y is the height and x is the horizontal distance the ball travels. We want to know if Marcuskicked the ball hard enough to go over the goal post, which is 10 feet high.
Mixed Review
28. Factor 6x2− x−12.29. Find the vertex of y =−1
4 x2−3x−12 by completing the square.30. Solve using the quadratic formula: −4x2−15 =−4x.31. How many centimeters are in four fathoms? (Hint: 1 fathom = 6 feet)
32. Graph the solution to
{3x+2y≤−4x− y >−3
.
33. How many ways can 3 toppings be chosen from 7 options?
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1.10 Modeling with Quadratic Functions
Here you’ll learn how to find the quadratic equation that fits to a data set.
On reduced-gravity flights, airplanes fly in large parabolic arcs. Determine the quadratic equation of best fit for thedata set below that represents the arc in which the airplane flies.
TABLE 1.5:
x(Time in Sec.) 0 20 33 45y (Altitude in 1000feet)
24 29 33 29
Source: Nasa.gov (http://www.nasa.gov/audience/forstudents/5-8/features/what-is-microgravity-58.html)
Watch This
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James Sousa: Ex: Quadratic Regression on the TI84 - Stopping Distance
Guidance
When finding the equation of a parabola, you can use any of the three forms. If you are given the vertex and anyother point, you only need two points to find the equation. However, if you are not given the vertex you must haveat least three points to find the equation of a parabola.
Example A
Find the equation of the parabola with vertex (-1, -4) and passes through (2, 8).
Solution: Use vertex form and substitute -1 for h and -4 for k.
y = a(x− (−1))2−4
y = a(x+1)2−4
Now, take the second point and plug it for x and y and solve for a.
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8 = a(2+4)2−4
12 = 36a13= a
The equation is y = 13(x+1)2−4.
Like in the Analyzing Scatterplots lesson, we can also fit a set of data to a quadratic equation. In this concept, wewill be using quadratic regression and a TI-83/84.
Example B
Determine the quadratic equation of best fit for the data set below.
TABLE 1.6:
x 0 4 7 12 17y 7 9 10 8 3
Solution: We need to enter the x−coordinates as a list of data and the y−coordinates as another list.
1. Press STAT.
2. In EDIT, select 1:Edit. . . . Press ENTER.
3. The List table appears. If there are any current lists, you will need to clear them. To do this, arrow up to L1 sothat it is highlighted (black). Press CLEAR, then ENTER. Repeat with L2, if necessary.
4. Now, enter the data into the lists. Enter all the entries into L1 (x) first and press enter between each entry. Then,repeat with L2 and y.
5. Press 2nd MODE (QUIT).
Now that we have everything in the lists, we can use quadratic regression to determine the equation of best fit.
6. press STAT and then arrow over to the CALC menu.
7. Select 5:QuadReg. Press ENTER.
8. You will be taken back to the main screen. Type (L1,L2) and press ENTER. L1 is 2nd 1, L2 is 2nd 2.
9. The following screen appears. The equation of best fit is y =−0.64x2 +0.86x+6.90.
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If you would like to plot the equation on the scatterplot follow the steps from the Finding the Equation of Best Fitusing a Graphing Calculator concept. The scatterplot and parabola are to the right.
This technique can be applied to real-life problems. You can also use technique to find the equation of any parabola,given three points.
Example C
Find the equation of the parabola that passes through (1, 11), (2, 20), (-3, 75).
Solution: You can use the same steps from Example B to find the equation of the parabola. Doing this, you shouldget the equation is y = 5x2−6x+12.
This problem can also be done by solving three equations, with three unknowns. If we plug in (x,y) to y = ax2 +bx+ c, we would get:
11 = a+b+ c
20 = 4a+2b+ c
75 = 9a−3b+ c
Use linear combinations to solve this system of equations (see Solving a System in Three Variables Using LinearCombinations concept). This problem will be finished in the Problem Set.
Intro Problem Revisit Use your calculator to find the quadratic equation of best fit for the given table.
TABLE 1.7:
x(Time in Sec.) 0 20 33 45y (Altitude in 1000feet)
24 29 33 29
y =−0.0081x2 +0.495x+23.6987 is the quadratic equation of best fit for the data.
Guided Practice
1. Find the equation of the parabola with x−intercepts (4, 0) and (-5, 0) that passes through (-3, 8).
2. A study compared the speed, x (in miles per hour), and the average fuel economy, y (in miles per gallon) of asports car. Here are the results.
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TABLE 1.8:
speed 30 40 50 55 60 65 70 80fueleconomy
11.9 16.1 21.1 22.2 25.0 26.1 25.5 23.2
Plot the scatterplot and use your calculator to find the equation of best fit.
Answers
1. Because we are given the intercepts, use intercept form to find the equation.
y = a(x−4)(x+5) Plug in (-3,8) and solve for a
8 = a(−3−4)(−3+5)
8 =−14a
−47= a
The equation of the parabola is y =−47(x−4)(x+5).
2. Plotting the points, we have:
Using the steps from Example B, the quadratic regression equation is y =−0.009x2 +1.24x−18.23.
Vocabulary
Quadratic RegressionThe process through which the equation of best fit is a quadratic equation.
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Practice
Find the equation of the parabola given the following points. No decimal answers.
1. vertex: (-1, 1) point: (1, -7)2. x−intercepts: -2, 2 point: (4, 3)3. vertex: (9, -4) point: (5, 12)4. x−intercepts: 8, -5 point: (3, 20)5. x−intercepts: -9, -7 point: (-3, 36)6. vertex: (6, 10) point: (2, -38)7. vertex: (-4, -15) point: (-10, 1)8. vertex: (0, 2) point: (-4, -12)9. x−intercepts: 3, 16 point: (7, 24)
Use a graphing calculator to find the quadratic equation (in standard form) that passes through the given three points.No decimal answers.
10. (-4, -51), (-1, -18), (4, -43)11. (-5, 131), (-1, -5), (3, 51)12. (-2, 9), (2, 13), (6, 41)13. Challenge Finish computing Example C using linear combinations.
For the quadratic modeling questions below, use a graphing calculator. Round any decimal answers to the nearesthundredth.
14. The surface of a speed bump is shaped like a parabola. Write a quadratic model for the surface of the speedbump shown.
15. Physics and Photography Connection Your physics teacher gives you a project to analyze parabolic motion.You know that when a person throws a football, the path is a parabola. Using your camera, you take an longexposure picture of a friend throwing a football. A sketch of the picture is below.
You put the path of the football over a grid, with the x−axis as the horizontal distance and the y−axis as theheight, both in 3 feet increments. The release point, or shoulder height, of your friend is 5 ft, 3 in and youestimate that the maximum height is 23 feet. Find the equation of the parabola.
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16. An independent study was done linking advertising to the purchase of an object. 400 households were used inthe survey and the commercial exposure was over a one week period. See the data set below.
TABLE 1.9:
# of timescommer-cial wasshown, x
1 7 14 21 28 35 42 49
# ofhouse-holdsboughtitem, y
2 25 96 138 88 37 8 6
a) Find the quadratic equation of best fit.
b) Why do you think the amount of homes that purchased the item went down after more exposure to the commercial?
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CHAPTER 2 Non-Linear RelationshipsChapter Outline
2.1 PIECEWISE FUNCTIONS
2.2 GRAPHING BASIC ABSOLUTE VALUE FUNCTIONS
2.3 USING THE GENERAL ABSOLUTE VALUE EQUATION AND THE GRAPHING CAL-CULATOR
2.4 GRAPHS OF SQUARE ROOT FUNCTIONS
2.5 SHIFTS OF SQUARE ROOT FUNCTIONS
2.6 GRAPHING CUBED ROOT FUNCTIONS
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2.1 Piecewise Functions
Here you will explore the process of modeling functions with cubed functions or with different functions in differentparts of the model (piece-wise functions). Many real-life situations do not easily fit completely within a simplemodel, so it can be valuable to understand the process of using multiple models in the same situation.
As we venture into non-linear functions, we need to consider: How do exponential, quadratic, and polynomialfunctions compare?
Check out this link:
http://www.shmoop.com/common-core-standards/ccss-hs-f-le-3.html
Read through the information and try the Sample Assignment.
Guidance
Piece-wise functions may be used to model the interactions of multiple items each previously modeled by a simplerfunction. There are two special cases of piece-wise functions: step functions and absolute value functions. Thissection will introduce you to general piece-wise functions as well as step functions. The next few sections willcover absolute value functions.
Example A
Piece-wise functions can be used to describe situations in which quantitative relationships are different in differentintervals within the domain of the function. For example, consider a situation in which a wireless provider offerscustomers a monthly plan that costs $50, but then charges $0.40 cents per minute for every minute over 1000 includeddaytime minutes.
Model the monthly cost, C, of the plan as a function of m, the number of daytime minutes you use:
Solution:
C(m) =
{50, m≤ 100050+0.40(m−1000), m > 1000
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This function is comprised of a constant function, and a linear function with slope 0.40. If in a given month youuse 1000 minutes or fewer, your monthly cost is a constant $50. If you use more than 1000, each additional minuteinfluences the value of C. For example, if you use 1,020 minutes, your cost is:
TABLE 2.1:
C(1020) = 50 + 0.40(1020 - 1000)= 50 + 0.4(20)= 50 + 8=$58.00
It is important to note that in this kind of situation, the time used may to be rounded to the nearest minute. So, forexample, if you use 20.5 minutes, you will be charged for 21 minutes. This is an example of a non-continuous, ordiscontinuous, function, where there are definite steps from value to value rather than a smooth line connecting allpossible values.
For additional videos on piece-wise functions please click here and here .
Example B
The graph of a step function is a series of line segments.One example of a step function is the Greatest IntegerFunction.
For examples on Step Functions and the Greatest Integer Function please click here and here.
Guided Practice
Questions
1) At your favorite gourmet store, you buy loose tea by the ounce. The store does not calculate for fractions of anounce. Instead, they round the weight of tea UP to the nearest ounce. Your favorite tea costs $3.00 per ounce. Sketcha function to model the cost of tea as a function of the number of ounces you purchase. Explain why the function is
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not continuous.
Solutions
1) The function is not continuous because the function is constant between each whole number value of the domain,but then it “jumps up” to the next whole number value in the range. For example, 1.99 ounces of tea and 2.00 ouncesboth cost $6.00, but 2.01 ounces costs $9.00.
Practice
For 1-6, graph each function.
1. f (x) = 12 [[x]]
2. h(x) = [[2x]]
3. f (x) ={
2 i f x < 04 i f x≥ 0
}4. f (x) =
{−x i f x < 0
x+1 i f x≥ 0
}5. f (x) =
{2x−1 i f x >−1−x i f x≤−1
}6. f (x) =
{−1 i f x < 1
2x−2 i f x≥ 1
}7. Nicole wants to take a taxi from a hotel to a restaurant. The rate is $4 plus $1.50 per mile after the first
mile. Every fraction of a mile is rounded up to the next mile. Draw a graph to represent the cost of using the
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taxi cab. What is the cost if the trip is 10.5 miles long?8. A basketball team is ordering jackets with the team logo on the front and the player’s names on the back. A
graphic design firm charges $12 to set up the art work plus $25 per jacket, $5 for the team logo, and $3 toprint the last name for an order of 10 jackets or less. For orders of 11-20 jackets, a 5% discount is given. Fororders of more than 20 jackets, a discount of 10% is given.
a. Organize the information into a table. Include a column showing the total order price for each size order.
b. Write an equation representing the total price for an order of x shirts.
c. Graph the piecewise relation.
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2.2 Graphing Basic Absolute Value Functions
Here you’ll learn about the basic properties of absolute value functions.
While on vacation, you go scuba diving. You start at an unknown sea level of zero feet or higher. You then dive toa depth of 90 feet below sea level. What is the vertex of the absolute value function that represents your possibledistance from sea level after diving?
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James Sousa: Ex: Graph an Absolute Value Function Using a Table of Values
Guidance
In the Solving Absolute Value Equations concept, we learned how to solve and define absolute value equations. Wewill now take this idea one step further and graph absolute value equations.
Investigation: Graphing the Parent Graph of an Absolute Value Function
1. We are going to graph y = |x|. Draw a table for x and y, with the x−values ranging from -3 to 3.
TABLE 2.2:
x |x| y−3 |−3| 3−2 |−2| 2−1 |−1| 10 |0| 01 |1| 12 |2| 23 |3| 3
2. Recall that the absolute value of a number is always positive. Now that you have 7 sets of points, plot each oneand graph the function.
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3. Notice that this function is very similar to the linear function, y = x. Draw this line on the graph in a differentcolor or with a dashed line.
4. Now, fold your graph on the x−axis. What do you notice?
In the investigation, you should discover that when you fold your graph on the x−axis, the line y = x becomes theabsolute value equation, y = |x|. That is because the absolute value of a number can never be zero; therefore, therange will always be positive. We call y = |x| the parent graph because it is the most basic of all the absolute valuefunctions. We will also compare other absolute value functions to this graph. All linear absolute value functionshave this “V” shape.
In general, we can define the graph of y = |x| as y =
{x; x≥ 0−x; x < 0
. From this, we see that each side, is the mirror
image of the other over a vertical line, through the vertex.
Example A
Use a table to graph y = |x−3|. Determine the domain and range.
Solution: In general, when you use a table to graph a function, pick some positive and negative numbers, as wellas zero. Use the equation to help you determine which x−values to pick. Setting what is inside the absolute valueequal to zero, we get that x = 3. Pick three values on either side of x = 3 and then graph.
TABLE 2.3:
x |x−3| y0 |−3| 31 |−2| 22 |−1| 13 |0| 04 |1| 15 |2| 26 |3| 3
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Notice that this graph shifts to the right 3 when compared to the parent graph. The domain will be all real numbers,x ∈ R , and the range will be all positive real numbers, including zero, y ∈ [0,∞). The domain and range can also beexpressed as (−∞,∞).
Example B
Use a table to graph y = |x|−5. Determine the domain and range.
Solution: Be careful! Here, the minus 5 is not inside the absolute value. So, first take the absolute value of thex−value and then subtract 5. In cases like these, the range can include negative numbers.
TABLE 2.4:
x |x|−5 y−3 |−3|−5 −2−2 |−2|−5 −3−1 |−1|−5 −40 |0|−5 −51 |1|−5 −42 |2|−5 −33 |3|−5 −2
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Here, the graph shifts down 5 when compared to the parent graph. The domain will be all real numbers, x ∈ R, andthe range will be all real numbers greater than or equal to -5, y ∈ [−5,∞). The domain can also be expressed as(−∞,∞) and the range as [−5,∞).
In these three absolute value graphs, you may have noticed that there is a minimum point. This point is called thevertex. For example, in Example B, the vertex is (0, -5). The vertex can also be a maximum. See the next example.
Example C
Use a table to graph y =−|x−1|+2. Determine the vertex, domain, and range.
Solution: Determine what makes the inside of the absolute value equation zero, x = 1. Then, to make your table ofvalues, pick a couple values on either side of x = 1.
TABLE 2.5:
x −|x−1|+2 y-2 −|−2−1|+2 -1-1 −|−1−1|+2 00 −|0−1|+2 11 −|1−1|+2 22 −|2−1|+2 13 −|3−1|+2 04 −|4−1|+2 -1
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The vertex is (1, 2) and in this case, it is the maximum value. The domain is x ∈ R, and the range is y ∈ (−∞,2].
Intro Problem Revisit The absolute value function that represents this situation is y = |x−90|, where x is your sealevel before diving. By graphing this function, you can see that the vertex occurs at the point (90, 0)
Vocabulary
Absolute ValueThe distance away from zero a number is. The absolute value is always positive.
Parent GraphThe simplest form of a particular type of function. All other functions of this type are usually compared to theparent graph.
VertexThe highest or lowest point of a graph.
MinimumThe lowest point of a graph. The minimum will yield the smallest value of the range.
MaximumThe highest point of a graph. The maximum will yield the largest value of the range.
Guided Practice
Graph the following functions using a table. Determine the vertex, domain, and range of each function.
1. y =−|x−5|
2. y = |x+4|−2
Answers
1. Determine what makes the inside of the absolute value equation zero, x = 5. Then, to make your table of values,pick a couple values on either side of x = 5.
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TABLE 2.6:
x −|x−5| y2 −|2−5| -33 −|3−5| -2-4 −|4−5| -15 −|5−5| 06 −|6−5| -17 −|7−5| -28 −|8−5| -3
The vertex is (5, 0) and in this case, it is the maximum value. The domain is x ∈ R, and the range is y ∈ (−∞,0].
2. Determine what makes the inside of the absolute value equation zero, x =−4. Then, to make your table of values,pick a couple values on either side of x =−4.
TABLE 2.7:
x |x+4|−2 y−1 |−1+4|−2 1-2 |−2+4|−2 0-3 |−3+4|−2 -1-4 |−4+4|−2 -2-5 |−5+4|−2 -1-6 |−6+4|−2 0-7 |−7+4|−2 1
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The vertex is (-4, -2) and in this case, it is the minimum value. The domain is x ∈ R, and the range is y ∈ [−2,∞).
Practice
Graph the following functions using a table. Determine the vertex, domain, and range of each function.
1. y = |x+6|2. y = |x−4|3. y =−|x|+34. y = |x|−25. y =−|x+3|+76. y = |x−1|−67. y = 2|x|8. y =−3|x|9. y = 1
3 |x|
Use problems 1-9 to answer fill in the blanks.
10. If there is a negative sign in front of the absolute value, the graph is ________________ (when compared tothe parent graph).
11. If the equation is y = |x−h|+k, the vertex will be ___________________.12. The domain of an absolute value function is always ____________________________.13. For y = a|x|, if a > 1, then the graph will be ___________________ than the parent graph.14. For y = a|x|, if 0 < a < 1, then the graph will be ___________________ than the parent graph.15. Without making a table, what is the vertex of y = |x−9|+7?16. The number of diners in a restaurant is modeled by d(t) =−0.5|t−288|+144 where t is the time (in minutes)
since the store opened at 10:00 A.M.
a. For what value of t are there 100 shoppers in the store?
b. At what time are there 100 shoppers in the store?
c. What is the greatest number of shoppers in the store?
d. At what time does the greatest number of shoppers occur?
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2.3 Using the General Absolute Value Equationand the Graphing Calculator
Here you’ll learn how to graph more complicated absolute value functions and use the graphing calculator.
Mrs. Patel assigns the absolute value function y =−|x+3|−2. She tells her students to find the vertex.
"This is hard," George laments. "I’m going to need a calculator."
"No, it’s not," Sarai counters. "I can tell you what the vertex is without even graphing by hand."
Who is right and what is the vertex?
Guidance
In the problem set of the previous concept, we were introduced to the general equation of an absolute value function.Let’s formally define it here.
General Form of an Absolute Value Function: For any absolute value function, the general form is y= a|x−h|+k,where a controls the width of the “V” and (h,k) is the vertex.
You probably made these connections during the problem set from the previous concept. Now, we will put it all touse together.
Example A
Graph y = |x|, y = 12 |x|, and y = 2|x| on the same set of axes. Compare the three functions.
Solution: You can make a table for all three of these functions. However, now that we have a better understandingof absolute value functions, let’s use some patterns. First, look at the vertex. Nothing is being added or subtracted,so the vertex for all three will be (0, 0). Second, look at “a.” For an absolute value function, we can think of a likethe slope. Referring back to the definition of the parent graph, each function above can be rewritten as:
y =
{x;x≥ 0−x;x < 0
(blue), y =
{12 x;x≥ 0−1
2 x;x < 0(red), and y =
{2x;x≥ 0−2x;x < 0
(green)
Comparing the three, we see that if the slope is between 1 and 0, the opening is wider than the parent graph. If theslope, or a, is greater than 1, then that opening is narrower. The amount of the opening between the two sides of anabsolute value function (and other functions) is called the breadth.
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Now, in addition to drawing a table, we can use the general form of an absolute value equation and the value of a tofind the shape of the V .
Example B
Without making a table, sketch the graph of y =−|x−6|−2.
Solution: First, determine the vertex. From the general form, we know that it will be (6, -2). Notice that the x−variable is the opposite sign of what is in the equation; the y−variable is the same. That is our starting point. Then,we have a negative sign in front of the absolute value. This means our V will open down. Finally, there is no a term,so we can assume it is 1, meaning that the slope of each side of the V will be 1 and -1.
Lastly, we can use a graphing calculator to help us graph absolute value equations. The directions given here pertainto the TI-83/84 series; however every graphing calculator should be able to graph absolute value functions.
Example C
Use a graphing calculator to graph y = |4x+1|−2. Find the vertex, domain, and range.
Solution: For the TI-83/84
1. Press the Y = button.
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2. Clear any previous functions (press CLEAR) and turn off any previous plots (arrow up to Plot 1 and pressENTER).
3. Press the MATH button, arrow over to NUM and highlight 1:abs(. Press ENTER.
4. Type in the remaining portion of the function. The screen:
5. Press GRAPH. If your screen is off, press ZOOM, scroll down to 6:ZStandard, and press ENTER.
The graph looks like:
As you can see from the graph, the vertex is not (-1, -2). The y−coordinate is -2, but the 4 inside the absolute valueaffects the x− coordinate. Set what is inside the absolute value equal to zero to solve for the x−coordinate of thevertex.
4x+1 = 0
4x =−1
x =−14
The vertex is(−1
4 ,−2). From the previous concept, we know that the domain is all real numbers. The range will
be any number greater than and including -2. In this function, the “a” term was inside the absolute value. When thishappens, it will always affect the x−coordinate of the vertex.
Intro Problem Revisit Sarai is right. The absolute value function is written in general form, so a calculator is notnecessary. The vertex is (–3, –2).
Vocabulary
General Form of an Absolute Value FunctionFor any absolute value function, the general form is y = a|x−h|+k, where a controls the width of the “V ” and(h,k) is the vertex.
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BreadthThe wideness or narrowness of a function with two symmetric sides.
Guided Practice
1. Graph y = 3|x+4|−5 without a graphing calculator or making a table. Find the vertex, domain, and range of thefunction.
2. Graph y =−2|x−5|+1 using a graphing calculator.
Answers
1. First, use the general form to find the vertex, (-4, -5). Then, use a to determine the breadth of the function. a = 3,so we will move up 3 and over 1 in both directions to find the points on either side of the vertex.
The domain is all real numbers and the range is all reals greater than and including -5.
Domain: x ∈ R or (−∞,∞)
Range: y ∈ [−5,∞) or [−5,∞)
2. Using the steps from Example C, the function looks like:
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Practice
1. Graph y = 3|x|, y =−3|x|, and y = |−3x| on the same set of axes. Compare the graphs.2. Graph y = 1
4 |x+1|, and y = 14 |x|+1 on the same set of axes. Compare the graphs.
3. Without graphing, do you think that y = 2|x|,y = |2x|, and y = |−2x| will all produce the same graph? Whyor why not?
4. We know that the domain of all absolute value functions is all real numbers. What would be a general rule forthe range?
Use the general form and pattern recognition to graph the following functions. Determine the vertex, domain, andrange. No graphing calculators!
5. y = |x−2|+56. y =−2|x+3|7. y = 1
3 |x|+48. y = 2|x+1|−29. y =−1
2 |x−7|10. y =−|x−8|+6
Use a graphing calculator to graph the following functions. Sketch a copy of the graph on your paper. Identify thevertex, domain, and range.
11. y =−4|2x+1|12. y = 2
3 |x−4|+12
13. y = 43 |2x−3|−7
2
Graphing Calculator Extension Use the graphing calculator to answer questions 14-16.
14. Graph y = x2−4 on your calculator. Sketch the graph and determine the domain and range.15. Graph y = |x2−4| on your calculator. Sketch graph and determine the domain and range.16. How do the two graphs compare? How are they different? What could you do to the first graph to get the
second?
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2.4 Graphs of Square Root Functions
Here you’ll learn how to graph and compare functions that involve square roots.
What if you had a square root function like y =√
2x+3. How would you graph that function? After completing thisConcept, you’ll be able to graph square root functions like this one and compare them to other square root functions.
Watch This
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CK-12 Foundation: Graphs of Square Root Functions
Guidance
In this chapter you’ll learn about a different kind of function called the square root function. You’ve seen that takingthe square root is very useful in solving quadratic equations. For example, to solve the equation x2 = 25 we take thesquare root of both sides:
√x2 =±
√25, so x =±5.
A square root function is any function with the form: y = a√
f (x)+ c —in other words, any function where anexpression in terms of x is found inside a square root sign (also called a “radical” sign), although other terms maybe included as well.
Graph and Compare Square Root Functions
When working with square root functions, you’ll have to consider the domain of the function before graphing.The domain is very important because the function is undefined when the expression inside the square root sign isnegative, and as a result there will be no graph in whatever region of x−values makes that true.
To discover how the graphs of square root functions behave, let’s make a table of values and plot the points.
Example A
Graph the function y =√
x.
Solution
Before we make a table of values, we need to find the domain of this square root function. The domain is found byrealizing that the function is only defined when the expression inside the square root is greater than or equal to zero.Since the expression inside the square root is just x, that means the domain is all values of x such that x≥ 0.
This means that when we make our table of values, we should pick values of x that are greater than or equal to zero.It is very useful to include zero itself as the first value in the table and also include many values greater than zero.This will help us in determining what the shape of the curve will be.
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TABLE 2.8:
x y =√
x0 y =
√0 = 0
1 y =√
1 = 12 y =
√2 = 1.4
3 y =√
3 = 1.74 y =
√4 = 2
5 y =√
5 = 2.26 y =
√6 = 2.4
7 y =√
7 = 2.68 y =
√8 = 2.8
9 y =√
9 = 3
Here is what the graph of this table looks like:
The graphs of square root functions are always curved. The curve above looks like half of a parabola lying on itsside, and in fact it is. It’s half of the parabola that you would get if you graphed the expression y2 = x. And the graphof y =−
√x is the other half of that parabola:
Notice that if we graph the two separate functions on the same coordinate axes, the combined graph is a parabolalying on its side.
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Now let’s compare square root functions that are multiples of each other.
Example B
Graph the functions y =√
x,y = 2√
x,y = 3√
x, and y = 4√
x on the same graph.
Solution
Here is just the graph without the table of values:
If we multiply the function by a constant bigger than one, the function increases faster the greater the constant is.
Example C
Graph the functions y =√
x,y =√
2x,y =√
3x, and y =√
4x on the same graph.
Solution
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Notice that multiplying the expression inside the square root by a constant has the same effect as multiplying by aconstant outside the square root; the function just increases at a slower rate because the entire function is effectivelymultiplied by the square root of the constant.
Also note that the graph of√
4x is the same as the graph of 2√
x. This makes sense algebraically since√
4 = 2.
Example D
Graph the functions y =√
x,y = 12
√x,y = 1
3
√x, and y = 1
4
√x on the same graph.
Solution
If we multiply the function by a constant between 0 and 1, the function increases more slowly the smaller the constantis.
Watch this video for help with the Examples above.
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CK-12 Foundation: Graphs of Square Root Functions
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Vocabulary
• A square root function is any function with the form: y = a√
f (x)+c —in other words, any function wherean expression in terms of x is found inside a square root sign (also called a “radical” sign).
Guided Practice
Graph the functions
a) y = 2√
x and y =−2√
x on the same graph.
b) y =√
x and y =√−x on the same graph.
Solutions:
a) If we multiply the whole function by -1, the graph is reflected about the x−axis.
b)
On the other hand, when just the x is multiplied by -1, the graph is reflected about the y−axis. Notice that thefunction y =
√−x has only negative x−values in its domain, because when x is negative, the expression under the
radical sign is positive.
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Practice
Graph the following functions.
1. y = 3√
x2. y =−1
2
√x
3. y =√
4x4. y =
√x+7
5. y = 2√
x−56. y =−
√3x+1−2
Graph the functions in problems 7-10 on the same coordinate axes.
7. y =√
x,y = 2.5√
x and y =−2.5√
x8. y =
√x,y = 0.3
√x and y = 0.6
√x
9. y =√
x,y =√
x−5 and y =√
x+510. y =
√x,y =
√x+8 and y =
√x−8
11. Describe the relationship between the graphs of y = x2 and y =√
x. How are they alike? How are theydifferent?
12. The time in seconds that it takes an object to fall a distance d is given by the function t = 14
√d (assuming zero
air resistance). Graph the function.13. The perimeter of a square is given by the function P = 4
√A, where A is the area of the square.
a. Graph the function.
b. Determine the perimeter of a square with an area of 729 m2.
c. Would the perimeter and area of a square every be the same value? Explain your reasoning.
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2.5 Shifts of Square Root Functions
Here you’ll learn what shifts result from performing operations both inside and outside the square root sign of squareroot functions. You’ll also learn how to graph such functions.
What if you had the square root function y =√
x? How would the graph of the function change if you added 5 to therighthand side of the equation or if you multiplied x by 3? After completing this Concept, you’ll be able to identifyvarious shifts in square root functions.
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CK-12 Foundation: Shifts of Square Root Functions
Guidance
We will now look at how graphs are shifted up and down in the Cartesian plane.
Example A
Graph the functions y =√
x,y =√
x+2 and y =√
x−2.
Solution
When we add a constant to the right-hand side of the equation, the graph keeps the same shape, but shifts up for apositive constant or down for a negative one.
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Example B
Graph the functions y =√
x,y =√
x−2, and y =√
x+2.
Solution
When we add a constant to the argument of the function (the part under the radical sign), the function shifts to theleft for a positive constant and to the right for a negative constant.
Now let’s see how to combine all of the above types of transformations.
Example C
Graph the function y = 2√
3x−1+2.
Solution
We can think of this function as a combination of shifts and stretches of the basic square root function y =√
x. Weknow that the graph of that function looks like this:
If we multiply the argument by 3 to obtain y =√
3x, this stretches the curve vertically because the value of yincreases faster by a factor of
√3.
Next, when we subtract 1 from the argument to obtain y =√
3x−1 this shifts the entire graph to the left by one unit.
Multiplying the function by a factor of 2 to obtain y = 2√
3x−1 stretches the curve vertically again, because yincreases faster by a factor of 2.
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Finally we add 2 to the function to obtain y = 2√
3x−1+2. This shifts the entire function vertically by 2 units.
Each step of this process is shown in the graph below. The purple line shows the final result.
Now we know how to graph square root functions without making a table of values. If we know what the basicfunction looks like, we can use shifts and stretches to transform the function and get to the desired result.
Watch this video for help with the Examples above.
MEDIAClick image to the left for more content.
CK-12 Foundation: Shifts of Square Root Functions
Vocabulary
• For the square root function with the form: y = a√
f (x)+ c, c is the vertical shift.
Guided Practice
Graph the function y =−√
x+3−5.
Solution
We can think of this function as a combination of shifts and stretches of the basic square root function y =√
x. Weknow that the graph of that function looks like this:
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Next, when we add 3 to the argument to obtain y =√
x+3 this shifts the entire graph to the right by 3 units.
Multiplying the function by -1 to obtain y =−√
x+3 which reflects the function across the x-axis.
Finally we subtract 5 from the function to obtain y = −√
x+3− 5. This shifts the entire function down verticallyby 5 units.
Practice
Graph the following functions.
1. y =√
2x−12. y =
√x−100
3. y =√
4x+44. y =
√5− x
5. y = 2√
x+56. y = 3−
√x
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7. y = 4+2√
x8. y = 2
√2x+3+1
9. y = 4+√
2− x10. y =
√x+1−
√4x−5
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2.6 Graphing Cubed Root Functions
Here you’ll graph a cubed root function with and without a calculator.
The next day, Mrs. Garcia assigns her student the cube root function y = − 3√(x+1) to graph for homework. The
following day, she asks her students which quadrant(s) their graph is in.
Alendro says that because of the negative sign, all y values are negative. Therefore his graph is only in the third andfourth quadrants quadrant.
Dako says that his graph is in the third and fourth quadrants as well but it is also in the second quadrant.
Marisha says they are both wrong and that her graph of the function is in all four quadrants.
Which one of them is correct?
Guidance
A cubed root function is different from that of a square root. Their general forms look very similar, y = a 3√x−h+kand the parent graph is y = 3√x. However, we can take the cubed root of a negative number, therefore, it will bedefined for all values of x. Graphing the parent graph, we have:
TABLE 2.9:
x y-27 -3-8 -2-1 -10 01 1
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TABLE 2.9: (continued)
x y8 227 3
For y = 3√x, the output is the same as the input of y = x3. The domain and range of y = 3√x are all real numbers.Notice there is no “starting point” like the square root functions, the (h,k) now refers to the point where the functionbends, called a point of inflection (see the Analyzing the Graph of a Polynomial Function concept).
Example A
Describe how to obtain the graph of y = 3√x+5 from y = 3√x.
Solution: From the previous concept, we know that the +5 indicates a vertical shift of 5 units up. Therefore, thisgraph will look exactly the same as the parent graph, shifted up five units.
Example B
Graph f (x) =− 3√x+2−3. Find the domain and range.
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Solution: From the previous example, we know that from the parent graph, this function is going to shift to the lefttwo units and down three units. The negative sign will result in a reflection.
Alternate Method: If you want to use a table (like in the previous concept), that will also work. Here is a table,then plot the points. (h,k) should always be the middle point in your table.
TABLE 2.10:
x y6 -5-1 -4-2 -3-3 -2-10 -1
Example C
Graph f (x) = 12
3√x−4.
Solution: The -4 tells us that, from the parent graph, the function will shift to the right four units. The 12 effects how
quickly the function will “grow”. Because it is less than one, it will grow slower than the parent graph.
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Using the graphing calculator: If you wanted to graph this function using the TI-83 or 84, press Y = and clear outany functions. Then, press (1÷ 2), MATH and scroll down to 4: 3
√ and press ENTER. Then, type in the rest ofthe function, so that Y =
(12
)3√
(X−4). Press GRAPH and adjust the window.
Important Note: The domain and range of all cubed root functions are both all real numbers.
Intro Problem Revisit If you graph the function y =− 3√
(x+1), you see that the domain is all real numbers, whichmakes all quadrants possible. However, for all positive values of x, y is negative because of the negative sign in frontof the cube root. That rules out the first quadrant. Therefore, Dako is correct.
Guided Practice
1. Evaluate y = 3√x+4−11 when x =−12.
2. Describe how to obtain the graph of y = 3√x+4−11 from y = 3√x.
Graph the following cubed root functions. Check your graphs on the graphing calculator.
3. y = 3√x−2−4
4. f (x) =−3√
x−1
Answers
1. Plug in x =−12 and solve for y.
y = 3√−12+4−11 =3√−8+4 =−2+4 = 2
2. Starting with y = 3√x, you would obtain y = 3√x+4−11 by shifting the function to the left four units and down11 units.
3. This function is a horizontal shift to the right two units and down four units.
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4. This function is a reflection of y = 3√x and stretched to be three times as large. Lastly, it is shifted down one unit.
Vocabulary
General Equation for a Cubed Root Functionf (x) = a 3√x−h+ k, where h is the horizontal shift and k is the vertical shift.
Practice
Evaluate f (x) = 3√2x−1 for the following values of x.
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1. f (14)2. f (−62)3. f (20)
Graph the following cubed root functions.
4. y = 3√x+45. y = 3√x−36. f (x) = 3√x+2−17. g(x) =− 3√x−68. f (x) = 2 3√x+19. h(x) =−3 3√x+5
10. y = 2 3√x+4−311. Describe the relationship between the graphs of y = x3 and y = 3√x. How are they alike? How are they
different?
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