Beginning Algebra, 11th Edition

We can generalize from Example 1.The x-coordinates of the x-intercepts for the

equation by the quadratic formula, are

and

Thus, the x-value of the vertex is half their sum.

Combine like terms.

Multiply; lowest terms

For the equation in Example 1, we have and

Thus, the x-value of the vertex is

which is the same x-value for the vertex we found in Example 1.(It can be shown

that the x-value of the vertex is x=- 2 ba , even if the graph has no x-intercepts.)

x=-

b

2 a

=-

- 2

2112

=1,

y=x^2 - 2 x-3, a=1, b=-2.

x

b

2 a

x=

1

2

a

• 2 b
  2 a

b

x=

1

2

a

• b+ 2 b^2 - 4 ac-b- 2 b^2 - 4 ac
  2 a

b

x=

1

2

a

• b+ 2 b^2 - 4 ac
  2 a

+

• b- 2 b^2 - 4 ac
  2 a

b

x=

• b- 2 b^2 - 4 ac
  2 a

x=.

• b+ 2 b^2 - 4 ac
  2 a

y=ax^2 +bx+c,

582 CHAPTER 9 Quadratic Equations

Graphing a Parabola

Graph

Step 1 Find the vertex. The x-value of the vertex is

The y-value of the vertex is

so the vertex is The axis is the line

Step 2 Now find the y-intercept. Let in.

The y-intercept is 10 , 12.

y= 02 - 4102 + 1 = 1

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

12 , - 32. x= 2.

y= 22 - 4122 + 1 = - 3 ,

x=- a=1,b=- 4

b

2 a

=-

- 4

2112

= 2.

y=x^2 - 4 x+1.

EXAMPLE 2

Graphing the Parabola

Step 1 Find the vertex.Let and find the corresponding y-value

by substituting for xin the equation.

Step 2 Find they-intercept.Let and solve for y.

Step 3 Find the x-intercepts(if they exist). Let and solve for x.

Step 4 Plotthe intercepts and the vertex.

Step 5 Find and plot additional ordered pairsnear the vertex and inter-

cepts as needed, using symmetry about the axis of the parabola.

y= 0

x= 0

x=- 2 ba ,

yax^2 bxc

http://www.ebook777.com

http://www.ebook777.com