5l2

Copyright McDougal Littell/Houghton Mifflin Company Lesson 5.2 Algebra 2 C&S Notetaking Guide 91

Graphing Quadratic Functions in Vertex or Intercept FormGoal p Graph quadratic functions in different forms.

VOCABULARY

Vertex form The form y 5 a(x 2 h)2 1 k, where the vertex of the graph is (h, k) and the axis of symmetry is x 5 h

Intercept form The form y 5 a(x 2 p)(x 2 q), where the x-intercepts of the graph are p and q

Minimum value The y-coordinate of the vertex for y 5 ax2 1 bx 1 c when a > 0

Maximum value The y-coordinate of the vertex for y 5 ax2 1 bx 1 c when a < 0

5.2

Your Notes

VERTEX FORM OF A QUADRATIC FUNCTION

STEPS FOR GRAPHING y 5 a(x 2 h)2 1 k

Step 1 Draw the axis of symmetry . It is the line x 5 h .

Step 2 Plot the vertex , ( h , k ).

Step 3 Plot two points on one side of the axis of symmetry . Use symmetry to plot two more points on the opposite side.

Step 4 Draw a parabola through the points.

n2nt-05-NTG.indd 91n2nt-05-NTG.indd 91 10/8/07 9:47:41 PM10/8/07 9:47:41 PM


INTERCEPT FORM OF A QUADRATIC FUNCTION

STEPS FOR GRAPHING y 5 a(x 2 p)(x 2 q)

Step 1 Draw the axis of symmetry . It is the line

x 5 p q

} 2 .

Step 2 Find and plot the vertex . The x -coordinate

of the vertex is p q

} 2 . Substitute this value for

x in the function to find the y -coordinate of the vertex.

Step 3 Plot the points where the x-intercepts , p and q, occur.

Step 4 Draw a parabola through the points.

Your Notes Example 1 Graph a Quadratic Function in Vertex Form

O

y

x1

1

Graph y 5 2(x 2 3)2 1 2.

Solution

The function is in vertex form y 5 (x 2 h)2 1 k where a 5 2 , h 5 3 , and k 5 2 . Because a > 0 , the parabola opens up .

1. Draw the axis of symmetry, x 5 h 5 3 .

2. Plot the vertex (h, k) 5 ( 3 , 2 ).

3. Plot points. The x-values 1 and 2 are to the right of the axis of symmetry.

x 5 1: y 5 2( 1 2 3)2 1 2 5 10

x 5 2: y 5 2( 2 2 3)2 1 2 5 4

Plot the points (1, 10 ) and (2, 4 ). Then plot their mirror images across the axis of symmetry.

4. Draw a parabola through the points.


Your Notes


Graph y 5 2(x 2 2)(x 2 4).

Solution

The function is in intercept form y 5 a(x 2 p)(x 2 q) where a 5 21 , p 5 2 , and q 5 4 . Because a < 0 , the parabola opens down .

1. Draw the axis of symmetry. The axis of symmetry is:

x 5 p 1 q

} 2 5 2 4

} 2 5 3

2. Find and plot the vertex. The x-coordinate of the vertex

is x 5 3 . Calculate the y-coordinate of the vertex.

y 5 2(x 2 2)(x 2 4)

5 2( 3 2 2)( 3 2 4) 5 1

Plot the vertex ( 3 , 1 ).

3. Plot the points where the x-intercepts occur. The x-intercepts are 2 and 4 . Plot the points ( 2 , 0) and ( 4 , 0).

4. Draw a parabola through the points.

Example 2 Graph a Quadratic Function in Intercept Form

O

y

x1

1

O

y

x1

1

x 5 4

(4, 22)

O

y

x1

1(21, 0)

(1, 24)

(3, 0)

1. Graph the function y 5 (x 2 4)2 2 2. Label the vertex and axis of symmetry.

2. Graph the function y 5 (x 2 3)(x 1 1). Label the vertex and the x-intercepts.

Checkpoint Complete the following exercises.

n2nt-05-NTG.indd 93n2nt-05-NTG.indd 93 10/29/07 11:16:16 AM10/29/07 11:16:16 AM

Your Notes

3. y 5 2(x 1 6)(x 2 2)

maximum; 16

4. y 5 3(x 2 1)(x 2 5)

minimum; 212

5. y 5 4(x 2 3)2 1 4

minimum; 4

Checkpoint Tell whether the function has a minimum or maximum value. Find the minimum or maximum value of the functions.

Homework


Example 3 Find the Minimum or Maximum Value

Tell whether the function y 5 2(x 1 4)(x 2 2) has a minimum value or a maximum value. Then find the minimum or maximum value.

Solution

The function is in intercept form y 5 a(x 2 p)(x 2 q) where a 5 2 , p 5 24 , and q 5 2 . Because a > 0, the function has a minimum value. Find the y-coordinate of the vertex.

x 5 p 1 q

} 2 5 24 1 2 } 2 5 21

y 5 2 ( 21 1 4)( 1 2 2) 5 218

The minimum value of the function is 218 .


5l2

Documents

Transcript of 5l2