april 3, 2014

Holt Algebra 1

9-3 Graphing Quadratic FunctionsApril 3, 2014

Today:

2 STAR Tests CompletedWarm-Up

Getting to Know the Quadratic Function: (how the a, b, & c values change the parabola)

Class Work

Holt Algebra 1

9-3 Graphing Quadratic Functions

x = 0

x = 1

(0, 2)

1. y = 4x2 – 72. y = x2 – 3x + 1

Find the axis of symmetry.

3. y = –2x2 + 4x + 3

(2, -12) 5. y = x2 + 4x + 5 6. y = -2x2 + 2x – 8

Find the vertex and state whether the graph opens up or down.

Warm-Up

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For a quadratic function in the form y = ax2 + bx + c, when x = 0, y = c. The y-intercept of a quadratic function is c

Finding the Y intercept

Find the vertex and the y-intercept

1. y = x2 – 2 y = x2 – 4x + 4 y = -2x2 – 6x - 3

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9-3 Graphing Quadratic FunctionsEffects of the a, b, & c values

With your graph paper, graph the function: y = x2

This is called the parent function. All other quadratic functions are simply transformations of the parent.

For the parent function f(x) = x2:

• The axis of symmetry is x = 0, or the y-axis.

• The vertex is (0, 0)

• The function has only one zero, 0.

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The value of a in a quadratic function determines not only the direction a parabola opens, but also the width of the parabola.

Effects of the a, b, & c values

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Example 1A: Comparing Widths of Parabolas

Order the functions from narrowest graph to widest.

f(x) = 3x2, g(x) = 0.5x2, h(x) = 1.5x2

f(x) = 3x2

h(x) = 1.5x2

g(x) = 0.5x2

The function with the narrowest graph has the greatest |a|.

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Holt Algebra 1


The value of c in a quadratic function determines not only the value of the y-intercept but also a vertical translation of the graph of f(x) = ax2 up or down the y-axis.

Holt Algebra 1

9-3 Graphing Quadratic FunctionsComparing Graphs of Quadratic Functions

Compare the graph of the function with the graph of f(x) = x2

opens downward and the graph of

f(x) = x2 opens upward.

• The graph of

is wider than the graph of

• The graph of

f(x) = x2.

Holt Algebra 1

9-3 Graphing Quadratic FunctionsCompare the graph of each the graph of f(x) = x2.

g(x) = –x2 – 4

• The graph of g(x) = –x2 – 4 opens downward and the graph of f(x) = x2 opens upward.

The vertex of g(x) = –x2 – 4

f(x) = x2 is (0, 0).

is translated 4 units down to (0, –3).

• The vertex of

• The axis of symmetry is the same.

Holt Algebra 1


Example 1: Graphing a Quadratic Function

Graph y = 3x2 – 6x + 1.

Step 1 Find the axis of symmetry.

= 1

The axis of symmetry is x = 1.

Simplify.

Use x = . Substitute 3

for a and –6 for b.

Step 2 Find the vertex.

y = 3x2 – 6x + 1

= 3(1)2 – 6(1) + 1= 3 – 6 + 1

= –2

The vertex is (1, –2).

The x-coordinate of the vertex

is 1. Substitute 1 for x.

Simplify.

The y-coordinate is –2.

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Example 1 Continued

Step 3 Find the y-intercept.

y = 3x2 – 6x + 1

y = 3x2 – 6x + 1

The y-intercept is 1; the graph passes through (0, 1).

Identify c.

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.

Since the axis of symmetry is x = 1, choose x-values less than 1.

Let x = –1.

y = 3(–1)2 – 6(–1) + 1

= 3 + 6 + 1

= 10

Let x = –2.

y = 3(–2)2 – 6(–2) + 1

= 12 + 12 + 1

= 25

Substitute

x-coordinates.

Simplify.

Two other points are (–1, 10) and (–2, 25).

Example 1 Continued

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Graph y = 3x2 – 6x + 1.

Step 5 Graph the axis of

symmetry, the vertex, the point

containing the y-intercept, and

two other points.

Step 6 Reflect the points

across the axis of

symmetry. Connect the

points with a smooth curve.

Example 1 Continued

x = 1(–2, 25)

(–1, 10)

(0, 1)(1, –2)

x = 1

(–1, 10)

(0, 1)

(1, –2)

(–2, 25)

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Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point.

Helpful Hint

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Example 2

Graph the quadratic function.

y = 2x2 + 6x + 2


Simplify.


for a and 6 for b.

The axis of symmetry is x .

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y = 2x2 + 6x + 2

Simplify.

Example 2 Continued

= 4 – 9 + 2

= –2

The x-coordinate of the vertex is

. Substitute for x.

The y-coordinate is .

The vertex is .

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y = 2x2 + 6x + 2

y = 2x2 + 6x + 2


Identify c.

Example 2 Continued

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept.

Let x = –1

y = 2(–1)2 + 6(–1) + 1

= 2 – 6 + 2

= –2

Let x = 1

y = 2(1)2 + 6(1) + 2

= 2 + 6 + 2

= 10

Substitute

x-coordinates.

Simplify.

Two other points are (–1, –2) and (1, 10).

Example 2 Continued

Since the axis of symmetry is x = –1 , choose x

values greater than –1 .

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two other points.


across the axis of

symmetry. Connect the

points with a smooth curve.

y = 2x2 + 6x + 2

Example 2 Continued

(–1, –2)

(1, 10)

(–1, –2)

(1, 10)

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Class Work:Graphing Quadratic Functions

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Example 3

Graph the quadratic function.

y + 6x = x2 + 9


Simplify.


for a and –6 for b.


= 3

y = x2 – 6x + 9 Rewrite in standard form.

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Simplify.

Example 3 Continued

= 9 – 18 + 9

= 0

The vertex is (3, 0).

The x-coordinate of the vertex is

3. Substitute 3 for x.

The y-coordinate is 0. .

y = x2 – 6x + 9

y = 32 – 6(3) + 9

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y = x2 – 6x + 9

y = x2 – 6x + 9


Identify c.

Example 3 Continued

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Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y- intercept.

Since the axis of symmetry is x = 3, choose x-values less than 3.

Let x = 2

y = 1(2)2 – 6(2) + 9

= 4 – 12 + 9

= 1

Let x = 1

y = 1(1)2 – 6(1) + 9

= 1 – 6 + 9

= 4

Substitute

x-coordinates.

Simplify.

Two other points are (2, 1) and (1, 4).

Example 3 Continued

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two other points.


across the axis of symmetry.

Connect the points with a

smooth curve.

y = x2 – 6x + 9

Example 3 Continued

x = 3

(3, 0)

(0, 9)

(2, 1)

(1, 4)

(0, 9)

(1, 4)

(2, 1)

x = 3

(3, 0)

Holt Algebra 1


Example 2: Application

The height in feet of a basketball that is thrown can be modeled by f(x) = –16x2

+ 32x, where x is the time in seconds after it is thrown. Find the basketball’s maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air.

Holt Algebra 1


Example 2 Continued

1 Understand the Problem

The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the ground.

• The function f(x) = –16x2 + 32x models the height of the basketball after xseconds.

List the important information:

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2 Make a Plan

Find the vertex of the graph because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex. The basketball will hit the ground when its height is 0, so find the zeros of the function. You can do this by graphing.

Example 2 Continued

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Solve3


Use x = . Substitute

–16 for a and 32 for b.

Simplify.


Example 2 Continued

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f(x) = –16x2 + 32x

= –16(1)2 + 32(1)

= –16(1) + 32

= –16 + 32

= 16

The vertex is (1, 16).

The x-coordinate of

the vertex is 1.

Substitute 1 for x.

Simplify.

The y-coordinate is 16.

Example 2 Continued

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Identify c.f(x) = –16x2 + 32x + 0


Example 2 Continued

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Step 4 Graph the axis of symmetry, the vertex, and the point containing the y-intercept. Then reflect the point across the axis of symmetry. Connect the points with a smooth curve.

(0, 0)

(1, 16)

(2, 0)

Example 2 Continued

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The vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet. The graph shows the zeros of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds.

Example 2 Continued

(0, 0)

(1, 16)

(2, 0)

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Look Back4

Check by substitution (1, 16) and (2, 0) into the function.

16 = 16

0 = 0

Example 2 Continued

16 = –16(1)2 + 32(1)?

16 = –16 + 32?

0 = –16(2)2 + 32(0)?

0 = –64 + 64?

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The vertex is the highest or lowest point on a parabola. Therefore, in the example, it gives the maximum height of the basketball.

Remember!

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Check It Out! Example 2

As Molly dives into her pool, her height in feet above the water can be modeled by the function f(x) = –16x2 + 24x, where x is the time in seconds after she begins diving. Find the maximum height of her dive and the time it takes Molly to reach this height. Then find how long it takes her to reach the pool.

Holt Algebra 1


1 Understand the Problem

The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the pool.

Check It Out! Example 2 Continued

List the important information:

• The function f(x) = –16x2 + 24x models the height of the dive after x seconds.

Holt Algebra 1


2 Make a Plan

Find the vertex of the graph because the maximum height of the dive and the time it takes to reach it are the coordinates of the vertex. The diver will hit the water when its height is 0, so find the zeros of the function. You can do this by graphing.


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Solve3


Use x = . Substitute

–16 for a and 24 for b.

Simplify.

The axis of symmetry is x = 0.75.


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f(x) = –16x2 + 24x

= –16(0.75)2 + 24(0.75)

= –16(0.5625) + 18

= –9 + 18

= 9

The vertex is (0.75, 9).

Simplify.

The y-coordinate is 9.

The x-coordinate of

the vertex is 0.75.

Substitute 0.75 for x.


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Identify c.f(x) = –16x2 + 24x + 0



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Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept.

Since the axis of symmetry is x =

0.75, choose an x-value that is less than

0.75.Let x = 0.5

f(x) = –16(0.5)2 + 24(0.5)

= –4 + 12

= 8 Another point is (0.5, 8).

Substitute 0.5 for x.

Simplify.


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Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve.

(1.5, 0)

(0.75, 9)

(0, 0)

(0.5, 8) (1, 8)


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The vertex is (0.75, 9). So at 0.75 seconds, Molly's dive has reached its maximum height of 9 feet. The graph shows the zeros of the function are 0 and 1.5. At 0 seconds the dive has not begun, and at 1.5 seconds she reaches the pool. Molly reaches the pool in 1.5 seconds.

(1.5, 0)

(0.75, 9)

(0, 0)

(0.5, 8) (1, 8)


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Look Back4

Check by substitution (0.75, 9) and (1.5, 0) into the function.

9 = 9


0 = 0

9 = –16(0.75)2 + 24(0.75)?

9 = –9 + 18?

0 = –16(1.5)2 + 24(1.5)?

0 = –36 + 36?

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Lesson Quiz

1. Graph y = –2x2 – 8x + 4.

2. The height in feet of a fireworks shell can be modeled by h(t) = –16t2 + 224t, where t is the time in seconds after it is fired. Find the maximum height of the shell, the time it takes to reach its maximum height, and length of time the shell is in the air.

784 ft; 7 s; 14 s

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april 3, 2014

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