Algebra unit 10.2

UNIT 10.2 QUADRATIC FUNCTIONSUNIT 10.2 QUADRATIC FUNCTIONS

Warm Up

Find the axis of symmetry.

1. y = 4x2 – 7 2. y = x2 – 3x + 1

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

Find the vertex.

5. y = x2 + 4x + 5 6. y = 3x2 + 2

7. y = 2x2 + 2x – 8

x = 0

x = 1

(–2, 1) (0, 2)

Graph a quadratic function in the form y = ax2 + bx + c.

Objective

Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form y = ax2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.

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.

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.

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

Substitutex-coordinates.

Simplify.

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

Example 1 Continued

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)

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

Check It Out! Example 1a Graph the quadratic function.

y = 2x2 + 6x + 2

Step 1 Find the axis of symmetry.

Simplify.

Use x = . Substitute 2

for a and 6 for b.

The axis of symmetry is x .

Step 2 Find the vertex.

y = 2x2 + 6x + 2

Simplify.

Check It Out! Example 1a Continued

= 4 – 9 + 2

= –2

The x-coordinate of the vertex is

. Substitute for

x.

The y-coordinate is .

The vertex is .

Step 3 Find the y-intercept.

y = 2x2 + 6x + 2

y = 2x2 + 6x + 2

The y-intercept is 2; the graph passes through (0, 2).

Identify c.

Check It Out! Example 1a Continued

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

Substitutex-coordinates.

Simplify.

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

Check It Out! Example 1a Continued

Since the axis of symmetry is x = –1 , choose x values greater than –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.

y = 2x2 + 6x + 2Check It Out! Example 1a Continued

(–1, –2)

(1, 10)

(–1, –2)

(1, 10)

Check It Out! Example 1b

Graph the quadratic function.

y + 6x = x2 + 9

Step 1 Find the axis of symmetry.

Simplify.

Use x = . Substitute 1

for a and –6 for b.

The axis of symmetry is x = 3.

= 3

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

Step 2 Find the vertex.

Simplify.

Check It Out! Example 1b 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

Step 3 Find the y-intercept.

y = x2 – 6x + 9

y = x2 – 6x + 9

The y-intercept is 9; the graph passes through (0, 9).

Identify c.

Check It Out! Example 1b Continued

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 = 2y = 1(2)2 – 6(2) + 9 = 4 – 12 + 9 = 1

Let x = 1 y = 1(1)2 – 6(1) + 9

= 1 – 6 + 9

= 4

Substitutex-coordinates.

Simplify.

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

Check It Out! Example 1b Continued

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.

y = x2 – 6x + 9

Check It Out! Example 1b Continued

x = 3

(3, 0)

(0, 9)

(2, 1)

(1, 4)

(0, 9)

(1, 4)

(2, 1)

x = 3

(3, 0)

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.

Example 2 Continued

11 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 x seconds.

List the important information:

22 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

Solve33

Step 1 Find the axis of symmetry.

Use x = . Substitute

–16 for a and 32 for b.

Simplify.

The axis of symmetry is x = 1.

Example 2 Continued

Step 2 Find the vertex.

f(x) = –16x2 + 32x

= –16(1)2 + 32(1)

= –16(1) + 32

= –16 + 32

= 16The 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

Step 3 Find the y-intercept.

Identify c.f(x) = –16x2 + 32x + 0

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

Example 2 Continued

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

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)

Look Back44

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?

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

Remember!

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.

11 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.

22 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.

Check It Out! Example 2 Continued

Solve33

Step 1 Find the axis of symmetry.

Use x = . Substitute

–16 for a and 24 for b.

Simplify.

The axis of symmetry is x = 0.75.

Check It Out! Example 2 Continued

Step 2 Find the vertex.

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.

Check It Out! Example 2 Continued

Step 3 Find the y-intercept.

Identify c.f(x) = –16x2 + 24x + 0

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

Check It Out! Example 2 Continued

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.

Check It Out! Example 2 Continued

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)

Check It Out! Example 2 Continued

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)

Check It Out! Example 2 Continued

Look Back44

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

9 = 9

Check It Out! Example 2 Continued

0 = 0

9 = –16(0.75)2 + 24(0.75)?

9 = –9 + 18?

0 = –16(1.5)2 + 24(1.5)?

0 = –36 + 36?

Lesson Quiz1. 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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