Section 11.1 Quadratic Functions and Their Graphs Copyright © 2013, 2009, and 2005 Pearson...

Section 11.1

Quadratic Functions and Their Graphs

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Graphs of Quadratic Functions

• Min-Max Applications

• Basic Transformations of Graphs

• More About Graphing Quadratic Functions (Optional)

The graph of any quadratic function is a parabola.

The vertex is the lowest point on the graph of a parabola that opens upward and the highest point on the graph of a parabola that opens downward.

The graph is symmetric with respect to the y-axis. In this case the y-axis is the axis of symmetry for the graph.

Example

Use the graph of the quadratic function to identify the vertex, axis of symmetry, and whether the parabola opens upward or downward.a. b.

Vertex (0, 2)Axis of symmetry: x = –2Open: up

Vertex (0, 4)Axis of symmetry: x = 0Open: down

Example

Find the vertex for the graph of Support your answer graphically.Solutiona = 2 and b = 8

Substitute into the equation to find the y-value.

2( ) 2 8 3.f x x x

82

2(2)x

2

bx

a

2( ) 2( 2) 8( 2) 3

8 16 3

11

f x

The vertex is (2, 11), which is supported by the graph.

Example

Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.

The vertex is (0, –2)axis of symmetry x = 0

2( ) 2f x x

x f(x) = x2 – 2

3 7

2 2

1 1

0 2

1 1

2 2

3 7

Example

Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.

The vertex is (2, 0)axis of symmetry x = 2

2( ) ( 2)g x x

x g(x) = (x – 2)2

0 4

1 1

2 0

3 1

4 4

Example

Identify the vertex, and the axis of symmetry on the graph, then graph.SolutionBegin by making a table of values.Plot the points and sketch a smooth curve.

The vertex is (2, 0)axis of symmetry x = 2

2( ) 2 3h x x x

x h(x) = x2 – 2x – 3

2 5

1 0

0 3

1 4

2 3

3 0

4 5

Example

Find the maximum y-value of the graph of

SolutionThe graph is a parabola that opens downward because a < 0. The highest point on the graph is the vertex.a = 1 and b = 2

2( ) 2 3.f x x x

( 2)1

2 2( 1)

bx

a

2( ) ( 1) 2( 1) 3

4

f x

Example

A baseball is hit into the air and its height h in feet after t seconds can be calculated by a. What is the height of the baseball when it is hit?b. Determine the maximum height of the baseball.Solutiona. The baseball is hit when t = 0.

b. The graph opens downward because a < 0. The maximum height occurs at the vertex. a = –16 and b = 64.

2( ) 16 64 2.h t t t

2( ) 16 64 2h t t t 2(0) 16(0) 64(0) 2

2

h

64 642

2 2( 16) 32

bx

a

Example (cont)

2( ) 16 64 2 h t t t

2(2) 16(2) 64(2) 2

66

h

The maximum height is 66 feet.

2( ) 16 64 2 h t t t

Basic Transformations of Graphs

The graph of y = ax2, a > 0.As a increases, the resulting parabola becomes narrower.When a > 0, the graph of y = ax2 never lies below thex-axis.

Example

Compare the graph of g(x) = –4x2 to the graph of f(x) = x2. Then graph both functions on the same coordinate axes.SolutionBoth graphs are parabolas.The graph of g opens downward and is narrower than the graph of f.

Section 11.2

Parabolas and Modeling

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Vertical and Horizontal Translations

• Vertex Form

• Modeling with Quadratic Functions (Optional)

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward with vertex (0, 0).

All three graphs have the same shape.y = x2

y = x2 + 1 shifted upward 1 unity = x2 – 2 shifted downward 2 units

Such shifts are called translations because they do not change the shape of the graph only its position

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2

y = (x – 1)2 Horizontal shift to the right 1 unit

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward with vertex (0, 0).y = x2

y = (x + 2)2 Horizontal shift to the left 2 units

Example

Sketch the graph of the equation and identify the vertex.

SolutionThe graph is similar to y = x2 except it has been translated 3 units down.

The vertex is (0, 3).

2 3y x

Example

Sketch the graph of the equation and identify the vertex.

SolutionThe graph is similar to y = x2 except it has been translated left 4 units.

The vertex is (4, 0).

2( 4)y x

Example

Sketch the graph of the equation and identify the vertex.

SolutionThe graph is similar to y = x2 except it has been translated down 2 units and right 1 unit.

The vertex is (1, 2).

2( 1) 2y x

Example

Compare the graph of y = f(x) to the graph of y = x2. Then sketch a graph of y = f(x) and y = x2 in the same xy-plane.

SolutionThe graph is translated to the right 2 units and upward 3 units.The vertex for f(x) is (2, 3) and the vertex of y = x2 is (0, 0).The graph opens upward and is wider.

21( ) ( 2) 3

4f x x

Example

Write the vertex form of the parabola with a = 3 and vertex (2, 1). Then express the equation in the form y = ax2 + bx + c.SolutionThe vertex form of the parabola is where the vertex is (h, k).a = 3, h = 2 and k = 1

To write the equation in y = ax2 + bx + c, do the following:

2( ) ,y a x h k

2)3( 12y x

2)3( 12y x 2( 4 43 1)y x x

23 12 12 1y x x 23 12 13y x x

Example

Write each equation in vertex form. Identify the vertex. a. b.Solutiona. Because , add and subtract 16 on the right.

2 8 13y x x 22 8 7y x x

2 28

162 2

b

2 8 13y x x

2 16 1 68 13y x x

24 3y x

The vertex is (4, 3).

Example (cont)

b. This equation is slightly different because the leading coefficient is 2 rather than 1. Start by factoring 2 from the first two terms on the right side.

22 8 7y x x

2 24

42 2

b

2

2

2 8 7

2( 4 ) 7

y x x

x x

2 42 4 74y x x

2 42 4 7 8y x x

22 2 1y x The vertex is ( 2, 1).

Section 11.3

Quadratic Equations

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Basics of Quadratic Equations

• The Square Root Property

• Completing the Square

• Solving an Equation for a Variable

• Applications of Quadratic Equations

Basics of Quadratic Equations

Any quadratic function f can be represented by f(x) = ax2 + bx + c with a 0.

Examples:

2 2 21( ) 2 1, ( ) 2 , and ( ) 2 1

3f x x g x x x h x x x

Basics

The different types of solutions to a quadratic equation.

Example

Solve each quadratic equation. Support your results numerically and graphically. a. b. c. Solutiona. Symbolic: Numerical: Graphical:

23 2 0x 2 9 6x x 2 2 8 0x x

The equation has no real solutions because x2 ≥ 0 for all real numbers x.

x y

1 5

0 2

1 5

2

2

2

3 2 0

3 2

2

3

x

x

x

Example (cont)

b. 2 9 6x x 2

2

9 6

6 9 0

( 3)( 3) 0

3 0 or 3 0

3 or 3

x x

x x

x x

x x

x x

The equation has one real solution.

x y

5 4

4 1

3 0

2 1

1 4

Example (cont)

c. 2 2 8 0x x 2 2 8 0

( 2)( 4) 0

2 0 or 4 0

2 or 4

x x

x x

x x

x x

The equation has two real solutions.

x y

4 0

2 8

1 9

0 8

2 0

The Square Root Property

The square root property is used to solve quadratic equations that have no x-terms.

Example

Solve each equation. a. b. c.Solutiona.

2 10x 225 16 0x 2( 3) 16x

2 10x

2 10x

10x

b. 225 16 0x 225 16x

2 16

25x

16

25x

4

5x

c. 2( 3) 16x

( 3) 16x

3 4x

3 4x

1 or 7x

Real World Connection

If an object is dropped from a height of h feet, its distance d above the ground after t seconds is given by

2( ) 16d t h t

Example

A toy falls 40 feet from a window. How long does the toy take to hit the ground? Solution 2( ) 16d t h t

2( ) 40 16d t t 240 16 0t

216 40t

2 40

16t

40

16t 4 10 2 10

416

10

2 1.6 sec.

Example

Find the term that should be added to to form a perfect square trinomial.

SolutionCoefficient of x-term is –8, so we let b = –8. To complete the square we divide by 2 and then square the result.

2 8x x

2 28

162 2

b

2 2168 ( 4)x x x

Example

Solve the equationSolutionWrite the equation in x2 + bx = d form.

2 8 13 0x x

2 28

162 2

b

2 8 13x x 2 168 3 61 1x x

2( 4) 3x

4 3x

4 3x

5.73 or 2.27x

Example

Solve the equationSolutionWrite the equation in x2 + bx = d form.

20 2 8 7x x

2 24

42 2

b

22 8 7x x

2 442

47

x x

2 1( 2)

2x

12

2x

2 74

2x x

22

2x

12

2x

22

2x

1.29 or 2.71x

Example

Solve the equation for the specified variable.Solution

216 for w x x

216w x

16

wx

4

wx

2

16

wx

Example

Use of the Internet in Western Europe has increased dramatically. The figure shows a scatter plot of online users in Western Europe, together with a graph of a function f that models the data. The function f is given by: where the output is in millions of users. In this formula x = 6 corresponds to 1996, x = 7 to 1997, and so on, until x = 12 represents 2002.a. Evaluate f(10) and interpret the result.b. Graph f and estimate the year when the number of Internet users reached 85 million.c. Solve part (b) numerically.

2( ) 0.976 4.643 0.238f x x x

Example (cont)

Solutiona. Evaluate f(10) and interpret the result.

Because x = 10 corresponds to 2000, there were about 51.4 million users in 2000.

2( ) 0.976 4.643 0.238f x x x

2( ) 0.976(10) 4.643(10) .0238

51.4

f x

Example (cont)

Solutionb. Graph f and estimate the year when the number of Internet users reached 85 million.

c. Solve part (b) numerically.

2( ) 0.976 4.643 0.238f x x x

Section 11.4

Quadratic Formula

Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

Objectives

• Solving Quadratic Equations

• The Discriminant

• Quadratic Equations Having Complex Solutions

, 0 and 1,xf x a a a

The solutions to ax2 + bx + c = 0 with a ≠ 0 are given by

QUADRATIC FORMULA

2 4.

2

b b acx

a

Example

Solve the equation 4x2 + 3x – 8 = 0. Support your results graphically.SolutionSymbolic SolutionLet a = 4, b = 3 and c = − 8.

2 4

2

b b acx

a

23 3 4 4 8

2 4x

3 137

8x

3 137

8x

or 3 137

8x

1.1x 1.8x or

Example (cont)

4x2 + 3x – 8 = 0Graphical Solution

Example

Solve the equation 3x2 − 6x + 3 = 0. Support your result graphically.SolutionLet a = 3, b = −6 and c = 3.

2 4

2

b b acx

a

26 6 4 3 3

2 3x

6 0

6x

1x

Example

Solve the equation 2x2 + 4x + 5 = 0. Support your result graphically.SolutionLet a = 2, b = 4 and c = 5.

2 4

2

b b acx

a

24 4 4 2 5

2 2x

4 24

4x

There are no real solutions

for this equation because

is not a real number.

24

, 0 and 1,xf x a a a

To determine the number of solutions to the quadratic equation ax2 + bx + c = 0, evaluate the discriminant b2 – 4ac.

1. If b2 – 4ac > 0, there are two real solutions.

2. If b2 – 4ac = 0, there is one real solution.

3. If b2 – 4ac < 0, there are no real solutions; there are two complex solutions.

THE DISCRIMINANT AND QUADRATIC

EQUATIONS

Example

Use the discriminant to determine the number of solutions to −2x2 + 5x = 3. Then solve the equation using the quadratic formula.Solution−2x2 + 5x − 3 = 0Let a = −2, b = 5 and c = −3.

Thus, there are two solutions.

b2 – 4ac

= (5)2 – 4(−2)(−3) = 1

2 4

2

b b acx

a

5 1

2 2x

4

4x

1x

or

6

4x

1.5x

, 0 and 1,xf x a a a

If k > 0, the solution to x2 + k = 0 are given by

THE EQUATION x2 + k = 0

.x i k

Example

Solve x2 + 17 = 0.

SolutionThe solutions are

17.i

17 or 17.x i i

Example

Solve 3x2 – 7x + 5 = 0. Write your answer in standard form: a + bi. SolutionLet a = 3, b = −7 and c = 5. 2 4

2

b b acx

a

7 11

6x

and

27 7 4 3 5

2 3x

7 11

6

ix

7 11

6 6x i

7 11

6 6x i

Example

Solve Write your answer in standard form: a + bi. SolutionBegin by adding 2x to each side of the equation and then multiply by 5 to clear fractions.

Let a = −2, b = 10 and c = −15.

223 2 .

5 x

x

22 10 15 0x x

Example (cont)

Let a = −2, b = 10 and c = −15.

223 2 .

5 x

x

2 4

2

b b acx

a

10 20

4x

210 10 4 2 15

2 2x

10 2 5

4

ix

5 5

2 2x i

Example

Solve by completing the square.SolutionAfter applying the distributive property, the equation becomes

Since b = −4 ,add to each side of the equation.

4 5x x

2 4 5.x x

242 4

2 4 4 5 4x x 2

2 1x 2 1x 2x i

2x i

The solutions are 2 + i and 2 − i.