Intermediate Algebra (11th edition)

Completing the Square to Find the Vertex

Find the vertex of the graph of

Because the -term has a coefficient other than 1, we factor that coefficient out

of the first two terms before completing the square.

Factor out.

Add and subtract 1

within the parentheses.

Now bring outside the parentheses. Be sure to multiply it by

Distributive property

Factor. Combine like terms.

The vertex is 1 1, 2 2. NOW TRY

ƒ 1 x 2 =- 31 x- 122 + 2

=- 31 x^2 - 2 x+ 12 + 3 - 1

= - 31 x^2 - 2 x+ 12 + 1 - 321 - 12 - 1

 1 3.

=- 31 x^2 - 2 x+ 1 - 12 - 1

C^12 1 -^22 D^2 =^1 -^122 =^1

= - 31 x^2 - 2 x 2 - 1 - 3

ƒ 1 x 2 =- 3 x^2 + 6 x- 1

x^2

ƒ 1 x 2 =- 3 x^2 + 6 x-1.

EXAMPLE 2 1 a 12

9.6 More about Parabolas and Their Applications

OBJECTIVES

More About Parabolas and Their Applications

9.6

1 Find the vertex of a

vertical parabola.

2 Graph a quadratic

function.

3 Use the discriminant

to find the number

of x-intercepts of a

parabola with a

vertical axis.

4 Use quadratic

functions to solve

problems involving

maximum or

minimum value.

5 Graph parabolas

with horizontal

axes.

OBJECTIVE 1 Find the vertex of a vertical parabola.When the equation of a

parabola is given in the form there are two ways to locate the

vertex.

1. Complete the square, as shown in Examples 1 and 2,or

2. Use a formula derived by completing the square, as shown in Example 3.

Completing the Square to Find the Vertex

Find the vertex of the graph of

We can express in the form by completing the square

on as in Section 9.1.The process is slightly different here because we want

to keep alone on one side of the equation. Instead of adding the appropriate num-

ber to each side, we add and subtractit on the right.

Group the variable terms.

Add and subtract 4.

Bring outside the parentheses.

Factor. Combine like terms.

The vertex of this parabola is 1 2, 1 2. NOW TRY

ƒ 1 x 2 = 1 x- 222 + 1

= 1 x^2 - 4 x+ 42 - 4 + 5 - 4

= 1 x^2 - 4 x+ 4 - 42 + 5

C^12 1 - 42 D^2 = 1 - 222 = 4

= 1 x^2 - 4 x 2 + 5

ƒ 1 x 2 =x^2 - 4 x+ 5

ƒ 1 x 2

x^2 - 4 x,

x^2 - 4 x+ 5 1 x-h 22 + k

ƒ 1 x 2 = x^2 - 4 x+5.

EXAMPLE 1 1 a 12

ƒ 1 x 2 = ax^2 + bx+ c,

This is equivalent

to adding 0.

This is a key step.

NOW TRY
EXERCISE 1
Find the vertex of the graph of

ƒ 1 x 2 =x^2 + 2 x-8.

NOW TRY ANSWERS

1. 1 - 1, - 92 2. 1 2, 6 2

NOW TRY
EXERCISE 2
Find the vertex of the graph of

ƒ 1 x 2 =- 4 x^2 + 16 x-10.