Beginning Algebra, 11th Edition

Step 3 Let to determine the x-intercepts. The equation is

which cannot be solved by factoring, so we use the quadratic formula.

Let ,

in the quadratic formula.

Simplify.

Factor.

x= 2  23 Divide out 2.

x=

2 A 2  23 B

2

x= 212 = 24 # 23 = 223

4  223

2

x=

4  212

2

a=1,b=- 4 c= 1

x=

- 1 - 42  21 - 422 - 4112112

2112

y= 0 0 =x^2 - 4 x+1,

SECTION 9.5 More on Graphing Quadratic Equations; Quadratic Functions 583

NOW TRY
EXERCISE 2
Graphy=-x^2 + 4 x+2.

NOW TRY ANSWER
2.

0 x

y

2

2

y = –x^2 + 4x+ 2

(2, 6)

Factor first.

Then divide out

the common

factor.

Using a calculator, we find that the x-intercepts are and to

the nearest tenth.

Steps 4 Plot the intercepts, vertex, and the additional points shown in the table.

and 5 Connect these points with a smooth curve. The graph is shown in FIGURE 4.

1 3.7, 0 2 1 0.3, 0 2

y

• 1
  0
  0.. 3
  2
  3.. 7
  4
  5

6

1

0

• 3
  0
  1
  6

x

245

1

6

0

(2,–3)

x

y

y x^2 – 4 x+ 1

(^2) + 3
2 – 3

• 2


• 3

FIGURE 4 NOW TRY

OBJECTIVE 2 Use a graph to determine the number of real solutions of a

quadratic equation.Using the vertical line test (Section 3.6),we see that the graph

of an equation of the form

is the graph of a function. A function defined by an equation of the form

is called a quadratic function.Thedomain(possible x-values) of a quadratic func-

tion is the set of all real numbers, or The range(the resulting y-values) can

be determined after the function is graphed. In Example 2,the domain is

and from FIGURE 4, we see that the range is

InExample 2,we found that the x-intercepts of the graph of

have x-values

and

This means that and are also the solutions of the

equation 0=x^2 - 4 x+1.

2 - 23 L0.3 2 + 23 L3.7

2 - 23 L0.3 2 + 23 L3.7.

1 wherey= 02

y=x^2 - 4 x+ 1

3 - 3,q 2.

1 - q,q 2 ,

1 - q,q 2.

ƒ 1 x 2 ax^2 bxc 1 a 02

y=ax^2 +bx+c