Intermediate Algebra (11th edition)

OBJECTIVE 2 Graph a quadratic function.We give a general approach.

SECTION 9.6 More About Parabolas and Their Applications 543

Graphing a Quadratic Function yfx

Step 1 Determine whether the graph opens up or down.If the

parabola opens up. If , it opens down.

Step 2 Find the vertex.Use the vertex formula or completing the square.

Step 3 Find any intercepts. To find the x-intercepts (if any), solve

To find the y-intercept, evaluate

Step 4 Complete the graph.Plot the points found so far. Find and plot

additional points as needed, using symmetry about the axis.

ƒ 1 x 2 =0. ƒ 102.

a 60

a 7 0,

 1 2

Graphing a Quadratic Function

Graph the quadratic function defined by

Step 1 From the equation, so the graph of the function opens up.

Step 2 The vertex, was found in Example 3by using the vertex formula.

Step 3 Find any intercepts. Since the vertex, is in quadrant IV and the

graph opens up, there will be two x-intercepts. Let and solve.

Let

Factor.

or Zero-factor property

or Solve each equation.

The x-intercepts are and Find the y-intercept by evaluating

Let

The y-intercept is

Step 4 Plot the points found so far and additional points as needed using symmetry

about the axis, x=^12 .The graph is shown in FIGURE 13.

1 0, - 62.

ƒ 102 =- 6

ƒ 102 = 02 - 0 - 6 x=0.

ƒ 1 x 2 = x^2 - x- 6

ƒ 102.

1 3, 0 2 1 - 2, 0 2.

x= 3 x=- 2

x- 3 = 0 x + 2 = 0

0 = 1 x- 321 x+ 22

0 = x^2 - x- 6 ƒ 1 x 2 =0.

ƒ 1 x 2 = x^2 - x- 6

ƒ 1 x 2 = 0

A

1

2 , -

25

4 B,

A

1

2 , -

25

4 B,

a= 1,

ƒ 1 x 2 = x^2 - x- 6.

EXAMPLE 4

y

• 2


• 1
  0

0

• 4


• 6

2

3

• 4
  0

x

0

(0, –6)

(–1, –4) (2, –4)

x

y

(–2, 0) (3, 0)

1

2

x^12

-^254

()

1

2 , –

25

4

f (x) x^2 – x – 6

FIGURE 13

Vertex:

Axis:

Domain:

Range: C-^254 , qB

1 - q, q 2

x=^12

A^12 , -^254 B

ƒ 1 x 2 =x^2 - x- 6

NOW TRY

NOW TRY
EXERCISE 4
Graph the quadratic function
defined by

Give the vertex, axis, domain,
and range.

ƒ 1 x 2 =x^2 + 2 x-3.

x

y

x = –1

(–3, 0)

(–1, – 4) (0, – 3)

(1, 0)

f(x) = x^2 + 2x – 3

0

NOW TRY ANSWER
4.

vertex: ; axis: ;

domain: ; range: 1 - q, q 2 3 - 4, q 2

1 - 1, - 42 x=- 1