Lesson 9.2: Graph Essential Question: How do you graph general quadratic functions? Common Core...

Lesson 9.2: Graph Essential Question: How do you graph general quadratic functions?

Common CoreCC.9-12.F.BF.3Graph linear and quadratic functions and show intercepts, maxima, and minima.

Warm-up:

3-18-13

cbxaxy 2

Evaluate the expressions: 1.

2. + 9 when x = 2

3. Martin is replacing a square patch of counter top. The area of the patch is represented by . What is the area of the patch if the side length is 2.5 inches?

9-2 Graph General Quadratic Functions

Homework:Part One: 638/3-11; 28-36

Part Two: 638/ 16-26 even; 40-41; 46-58 even

a>0 a<0 a = 7 b = 2 c = 11

The graph opens upward.

a = -3 b = 9 c = 4

The graph opens downward.

Does the vertex contain a minimum or a maximum value?

a>0 y value in vertex is a MINIMUM

a<0 y value in vertex is a MAXIMUM

Tell whether the graph has a maximum value or minimum value.

Tell whether the graph opens up or down and whether it has a maximum value or minimum value.

Upward.Minimum.

Downward.Maximum.

Upward.Minimum.

Tell whether the graph opens up or down and whether it has a maximum value or minimum value.

Upward.Minimum.

Downward.Maximum.

Upward.Minimum.

• Axis of symmetry is the line:

2bxa

The x-coordinate of the vertex is on this line.

To find the y-coordinate of the vertex, substitute the value of x into the original equation.

Vertex:

Axis of Symmetry:

2bxa

( 4)2(1)

x

2x

2x

2 4 5y x x

2(2) 2 5)4(y

4 8 5y

1y

(2,1)2 4 5y x x

This is the Minimum.

Vertex:

Axis of Symmetry: 2

bxa

72( 1)

x

72

x

2 7 2y x x

2 7 27 7( ) ( )2 2

y

49 49 24 2

y

(72

, 414

)2 7 2y x x

49 98 84 4 4

y 414

y

72

x

This is the Maximum.

downup

><

c0 c

2bxa

2ba

2ba

Axis of Symmetry: 2

bxa

22(8)

x 1

8x

82(2)

x 2x 8

12( )2

x 8x

2bxa

Graph y = -2x2 + 4x + 1Step 1: Find the axis of symmetry.

42( 2)

x

x = 1

Step 2: Find the y-coordinate of the vertex.

Substitute the value of x into the original equation.

y = -2(1)2 + 4(1) + 1y = 3

Step 3: Find the y-intercept.

Substitute 0 for x in the original equation.y = -2(0)2 + 4(0) + 1y = 1

Step 4: Choose another value for x on the same side of the vertex.

Substitute -1 for x in the original equation.y = -2(-1)2 + 4(-1) + 1

y = -5

Step 5: Reflect the points across the axis of symmetry and draw the parabola.

2bxa

Graph y = x2 – 6x + 9Step One: Find the axis of symmetry.

Step Two: Find the y-coordinate of the vertex.Substitute the value of x into the original equation.

Step Three: Find the y-intercept.Substitute 0 for x in the original equation.

Step Four: Choose another value for x on the same side of the vertex.Substitute for x in the original equation.

Step Five: Reflect the points across the axis and draw the parabola.

y

12-10 -6 -2 4 8 12

12

-10

-6

-2

4

8

12

-12 -4 6

-12

-8

2

10

-8 10

-4

-12

-12

2

6

3x

2(3) 6(3) 9y 0y

9 so y-intercept is 9c

2(1) 6(1) 9y 4y

2bxa

2

2

(2) 4(2) 73

(2,3)

x

yy

2

1

(1) 4(1) 74

(1,4)

x

yy

vertex (2,3)

axis of symmetry 2x

This function has a minimum. The minimum value is y = 3.

2x

Upward.Minimum.

Downward.Maximum. Upward.

Minimum.

02(1)

x 0x

2(0) (0) 7f

(0) 7f

7 is the minimum value

02( 1)

x

0x

2(0) (0) 9f (0) 9f

9 is the maximum value

(4)2(2)

x 1x

2( 1) 2( 1) 4( 1)f

( 1) 2f

-2 is the minimum value

Find the vertex using 2

bxa

These are all y values of the vertex.

EXAMPLE 3 Find the minimum or maximum value

Tell whether the function f(x) = – 3x2 – 12x + 10 has aminimum value or a maximum value. Then find theminimum or maximum value.

SOLUTION

Because a = – 3 and – 3 < 0, the parabola opens down andthe function has a maximum value. To find the maximumvalue, find the vertex.

x = – = – = – 2 b2a

– 122(– 3)

f(–2) = – 3(–2)2 – 12(–2) + 10 = 22 Substitute –2 for x. Thensimplify.

The x-coordinate is – b2a

Find the minimum or maximum valueEXAMPLE 3

ANSWER

The maximum value of the function is f(– 2) = 22.

Find the minimum value of a functionEXAMPLE 4

The suspension cables between the two towers of the Mackinac Bridge in Michigan form a parabola that can be modeled by the graph of y = 0.000097x2 – 0.37x + 549 where x and y are measured in feet. What is the height of the cable above the water at its lowest point?

SUSPENSION BRIDGES

Find the minimum value of a functionEXAMPLE 4

SOLUTION

The lowest point of the cable is at the vertex of theparabola. Find the x-coordinate of the vertex. Use a = 0.000097 and b = – 0.37.

x = – = – ≈ 1910 b2a

– 0.372(0.000097)

Use a calculator.

Substitute 1910 for x in the equation to find they-coordinate of the vertex.

y ≈ 0.000097(1910)2 – 0.37(1910) + 549 ≈ 196

EXAMPLE 4

ANSWER

The cable is about 196 feet above the water at its lowest point.

Find the minimum value of a function

GUIDED PRACTICE for Examples 3 and 4

3. Tell whether the function f(x) = 6x2 + 18x + 13 has aminimum value or a maximum value. Then find theminimum or maximum value.

1 2

Minimum value;

ANSWER

GUIDED PRACTICE for Examples 3 and 4

SUSPENSION BRIDGES

4. The cables between the two towers of the Takoma Narrows Bridge form a parabola that can be modeled by the graph of the equation y = 0.00014x2 – 0.4x + 507 where x and y are measured in feet. What is the height of the cable above the water at its lowest point? Round your answer to the nearest foot.

ANSWER

221 feet

The highest point occurs at the Maximum.

The y value of the vertex is 10.

The height of the dome is 10 feet.

Summary

Classwork/Homework

• 9.2 Practice B worksheet