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Copyright © Cengage Learning. All rights reserved.
Quadratic Equations, Quadratic Functions, and Complex Numbers 9
Copyright © Cengage Learning. All rights reserved.
Section 9.49.4
Graphing Quadratic Functions
3
Objectives
Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex.
1. Find the vertex of a parabola by completing the
square.
Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c.
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22
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4
Objectives
5. Solve an application involving a quadratic equation.
6. Solve a quadratic equation using a graphing calculator.
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55
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Graphing Quadratic Functions
The function defined by the equation f (x) = mx + b is a linear function, because its right side is a first-degree polynomial in the variable x.
The function defined by f (x) = ax2 + bx + c (a ≠ 0) is a quadratic function, because its right side is asecond-degree polynomial in the variable x. In this section, we will discuss many quadratic functions.
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Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex1.
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Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex
A basic quadratic function, is defined by the equationf (x) = x2. Recall that to graph this function, we find several ordered pairs (x, y) that satisfy the equation, plot the pairs, and join the points with a smooth curve. A table of values and the graph appear in Figure 9-3.
Figure 9-3
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Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex
The graph of a quadratic function is called a parabola.
The lowest point (or minimum point) on the parabola that opens upward is called its vertex.
The vertex of the parabola shown in Figure 9-3 is the point V(0, 0).
If a parabola opens downward, its highest point (or maximum point) is the vertex.
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Example
Graph f (x) = x2 – 3. Compare the graph with Figure 9-3.
Figure 9-3
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Example – Solution
To find ordered pairs (x, y) that satisfy the equation, we select several numbers for x and compute the corresponding values of y. Recall that f (x) = y. If we let x = 3, we have
y = x2 – 3
y = 32 – 3
y = 6
Substitute 3 for x.
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Example – Solution
The ordered pair (3, 6) and others satisfying the equation appear in the table shown in Figure 9-4. To graph the function, we plot the points and draw a smooth curve passing through them.
Figure 9-4
cont’d
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Example – Solution
The resulting parabola is the graph of f (x) = x2 – 3. The vertex of the parabola is the point V(0, –3).
Note that the graph of f (x) = x2 – 3 looks just like the graph of f (x) = x2, except that it is 3 units lower.
cont’d
13
Graph a quadratic function of the form f (x) = ax2 + bx + c using a table of values and identify the vertex
Graphs of Parabolas
The graph of the function f (x) = ax2 + bx + c (a 0) is a parabola. It opens upward when a > 0, and it opens downward when a < 0.
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Find the vertex of a parabola by completing the square
2.
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Find the vertex of a parabola by completing the square
It is easier to graph a parabola when we know the coordinates of its vertex. We can find the coordinates of the vertex of the graph of
f (x) = x2 – 6x + 8
if we complete the square in the following way.
f (x) = x2 – 6x + 9 – 9 + 8
f (x) = (x – 3)2 – 1
Add 9 to complete the square on x2 –6x and then subtract 9.
Factor x2 – 6x + 9 and combine like terms.
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Find the vertex of a parabola by completing the square
Since a > 0 in the original equation, the graph will be a parabola that opens upward. The vertex will be the minimum point on the parabola, and the y-coordinate of the vertex will be the smallest possible value of y.
Because (x – 3)2 0, the smallest value of y occurs when (x – 3)2 = 0 or when x = 3. To find the corresponding value of y, we substitute 3 for x in the equation f (x) = (x – 3)2 – 1 and simplify.
f (x) = (x – 3)2 – 1
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Find the vertex of a parabola by completing the square
f (3) = (3 – 3)2 – 1
f (3) = 02 – 1
f (3) = –1
The vertex of the parabola is thepoint V(3, –1). The graph appearsin Figure 9-8.
Substitute 3 for x.
Figure 9-8
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Find the vertex of a parabola by completing the square
A generalization of this discussion leads to the following fact.
Graphs of Parabolas with Vertex at (h, k)
The graph of an equation of the form
f (x) = a(x – h)2 + k
is a parabola with its vertex at the point with coordinates(h, k). The parabola opens upward if a > 0, and it opens downward if a < 0.
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Example
Find the vertex of the parabola determined by f (x) = 2x2 + 8x + 2 and graph the parabola.
Solution:To find the vertex of the parabola, we will writef (x) = 2x2 + 8x + 2 in the form f (x) = a(x – h)2 + k by completing the square on the right side of the equation.
As a first step, we will make the coefficient of x2 equal to 1 by factoring 2 out of the binomial 2x2 + 8x. Then, we proceed as:
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Example – Solution
f (x) = 2x2 + 8x + 2
= 2(x2 + 4x) + 2
= 2(x2 + 4x + 4 – 4) + 2
= 2[(x + 2)2 – 4] + 2
= 2(x + 2)2 + 2(–4) + 2
= 2(x + 2)2 – 6
Or
f (x) = 2[x–2 (–2)]2 + (–6)
cont’d
Factor 2 out of 2x2 + 8x.
Complete the square on x2 + 4x.
Factor x2 + 4x + 4.
Distribute the multiplication by 2.
Simplify and combine like terms.
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Example – Solution
Since h = –2 and k = –6, the vertex of the parabola is the point V(–2, –6). Since a = 2, the parabola opens upward. In this case, the vertex will be the minimum point on the graph.
We can select numbers on either side of x = –2 to construct the table shown in Figure 9-10.
cont’d
Figure 9-10
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Example – Solution
To find the y-intercept, we substitute 0 for x in our original equation and solve for y: When x = 0, y = 2. Thus, the y-intercept is (0, 2).
We determine more ordered pairs, plot the points, and draw the parabola.
cont’d
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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c
3.
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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c
Much can be determined about the graph off (x) = ax2 + bx + c from the coefficients a, b, and c. We summarize these results as follows.
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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c
Graphing the Parabola f (x) = ax2 + bx + c
1. If a > 0, the parabola opens upward and the vertex is the minimum. If a < 0, the parabola opens downward and the vertex is the maximum.
2. The coordinates of the vertex are
3. The axis of symmetry is the vertical line
4. The y-intercept is (0, c).
5. The x-intercepts (if any) are determined by the solutions of ax2 + bx + c = 0.
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Example
Graph: f (x) = x2 – 2x – 3.
Solution:
The equation is in the form f (x) = ax2 + bx + c, with a = 1, b = –2, and c = –3.
Since a > 0, the parabola opens upward. To find the x-coordinate of the vertex, we substitute the values for a and b into the formula x = .
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Example – Solution
The x-coordinate of the vertex is x = 1. This is also the equation for the axis of symmetry. To find the y-coordinate,
we can find = f (1) by substituting 1 for x in the
equation and solving for y.
f (x) = x2 – 2x – 3
f (1) = 12 – 2 1 – 3
= 1 – 2 – 3
= –4
The vertex of the parabola is the point (1, –4).
cont’d
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Example – Solution
To graph the parabola, identify several other points with coordinates that satisfy the equation.
One easy point to find is the y-intercept. It is the value of y when x = 0. Thus, the parabola passes through the point (0, –3).
To find the x-intercepts of the graph, we set f (x) equal to 0 and solve the resulting quadratic equation:
f (x) = x2 – 2x – 3
0 = x2 – 2x – 3
cont’d
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Example – Solution
0 = (x – 3)(x + 1)
x – 3 = 0 or x + 1 = 0
x = 3 x = –1
Since the x-intercepts of the graph are (3, 0) and (–1, 0), the graph passes through these points.
Factor.
Set each factor equal to 0.
cont’d
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Example – Solution
The graph appears in Figure 9-11.
Figure 9-11
cont’d
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Identify the x-intercept, the y-intercept, the axis of symmetry, and the vertex of a parabola given a function in the form f (x) = ax2 + bx + c
Comment
If the entire parabola is above or below the x-axis, there will be no x-intercepts.
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Solve an application involving a quadratic equation4.
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Example – Finding Maximum Revenue
An electronics firm manufactures a high-quality smartphone. Over the past 10 years, the firm has learned
that it can sell x smartphones at a price of
dollars. How many smartphones should the firm manufacture and sell to maximize its revenue? Find the maximum revenue.
Solution:
The revenue obtained is the product of the number of smartphones that the firm sells (x) and the price of each
smartphone .
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Example – Solution
Thus, the revenue R is given by the formula
Since the graph of this function is a parabola that opens downward, the maximum value of R will be the value of R determined by the vertex of the parabola. Because the
x-coordinate of the vertex is at x = , we have
cont’d
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Example – Solution
If the firm manufactures 500 smartphones, the maximum revenue will be
cont’d
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Example – Solution
= 50,000
The firm should manufacture 500 smartphones to get a maximum revenue of $50,000.
cont’d
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Solve a quadratic equation using a graphing calculator
5.
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Solve a quadratic equation using a graphing calculator
We can use graphing methods to solve quadratic equations. For example, the solutions of the equation x2 – x – 3 = 0 are the values of x that will make y = 0 in the quadratic function f (x) = x2 – x – 3. To approximate these values, we graph the quadratic function and identify the x-intercepts.
If we use window values of x = [–10, 10] and y = [–10, 10] and graph the function f (x) = x2 – x – 3, using a TI84 graphing calculator wewill obtain the graph shown inFigure 9-12(a). Figure 9-12(a)
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Solve a quadratic equation using a graphing calculator
We can find the x-intercept exactly(if they are rational) by using theZERO command found in the CALCmenu. Enter values to the left andright of each x-intercept, similar to thesteps you followed to find theminimum/maximum.
In this case the x-intercepts are irrational. To find the exact values, we would have to use the quadratic formula.
Figure 9-12(b)