Beginning Algebra, 11th Edition

OBJECTIVE 1 Graph quadratic equations of the form

Every equation of the form

with has a graph that is a parabola. The vertex is an important point to locate

when graphing a quadratic equation.

aZ0,

y=ax^2 +bx+c,

1 a 02.

yax^2 bxc

SECTION 9.5 More on Graphing Quadratic Equations; Quadratic Functions 581

y

• 2


• 1
  0
  1
  2
  3
  4
  
  
  • 3
  
  
  • 4
  
  
  
  

5

0

• 3


• 4


• 3
  0
  5

x

3

1

5

0

(1, –4)

• 2 x

y

x 1

y x^2 – 2 x – 3

x-intercept

x-intercept

y-intercept

vertex

FIGURE 3

Graphing a Parabola by Finding the Vertex and Intercepts

Graph

We must find the vertex of the graph. Because of its symmetry, if a parabola has

two x-intercepts, the x-value of the vertex is exactly halfway between them.There-

fore, we begin by finding the x-intercepts. Let in the equation and solve for x.

Interchange sides.

Factor.

or Zero-factor property

or Solve each equation.

There are two x-intercepts, and

Since the x-value of the vertex is halfway between the x-values of the two

x-intercepts, it is half their sum.

x-value of the vertex

Find the corresponding y-value by substituting 1 for xin.

y-value of the vertex

The vertex is The axis is the line

To find the y-intercept, substitute in the equation.

The y-intercept is

Plot the three intercepts and the vertex. Find additional ordered pairs as needed.

For example, if then

leading to the ordered pair A table with all these ordered pairs is shown with

the graph in FIGURE 3.

1 2, - 32.

y= 22 - 2122 - 3 = - 3 ,

x=2,

10 , - 32.

y= 02 - 2102 - 3 =- 3

x= 0

11 , - 42. x=1.

y= 12 - 2112 - 3 = - 4

y=x^2 - 2 x- 3

x=

1

2

1 - 1 + 32 = 1

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

x=- 1 x= 3

x+ 1 = 0 x - 3 = 0

1 x+ 121 x- 32 = 0

x^2 - 2 x- 3 = 0

0 =x^2 - 2 x- 3

y= 0

y=x^2 - 2 x-3.

EXAMPLE 1

NOW TRY

NOW TRY
EXERCISE 1
Graph y=x^2 - x-2.

NOW TRY ANSWER

1. 
   

0 x

 , –^1294 

y

–1–2 2

y = x^2 – x – 2