Intermediate Algebra (11th edition)

NOW TRY

NOTE The vertical line is the y-axis.

Graphing a Line That Passes through the Origin

Graph x+ 2 y= 0.

EXAMPLE 5

x= 0

SECTION 3.1 The Rectangular Coordinate System 141

xy

0

• 1 5


• 1


• 1 - 4

x

y

0

(–1, 0)

Vertical

x + 1 = 0 line

FIGURE 8

To graph x 10

(or x 1), do not

simply graph the

point 1, 0. The

graph is a line.

1 - 2

=-

+ =

Find the x-intercept.

Let

Multiply.

x-intercept is

Find the y-intercept.

Let

Add.

y= 0 y-intercept is 1 0, 0 2.

2 y= 0

0 + 2 y= 0 x=0.

x+ 2 y= 0

x= 0 1 0, 0 2.

x+ 0 = 0

x+ 2102 = 0 y=0.

x+ 2 y= 0

Both intercepts are the same point, which means that the graph passes through

the origin. To find another point, choose any nonzero number for xor yand solve for

the other variable. We choose

Let

Subtract 4.

Divide by 2.

This gives the ordered pair As a check, verify that also lies on the

line. The graph is shown in FIGURE 9.

1 4, - 22. 1 - 2, 1 2

y=- 2

2 y=- 4

4 + 2 y= 0 x=4.

x+ 2 y= 0

x= 4.

1 0, 0 2 ,

x

y

0

(0, 0)

x-intercept

and

y-intercept

x + 2y = 0

(–2, 1)

(4, –2)

FIGURE 9

xy

1

00

4 - 2

• 2

NOW TRY

NOW TRY
EXERCISE 4
Graph x+ 3 =0.

NOW TRY ANSWERS
4.

5. 
   

–3

3

x + 3 = 0

0

y

x

–3 2

NOW TRY

EXERCISE 5

Graph. 2 x+ 3 y= 0

–2

2

0

y

x

–3 3

2 x + 3y = 0

Graphing a Vertical Line

Graph

The form shows that every value of yleads to making

the x-intercept No value of ymakes so the graph has no y-intercept.

A straight line that has no y-intercept is vertical. See FIGURE 8.

1 - 1, 0 2. x=0,

1 x+ 0 y=- 1 x=-1,

x+ 1 = 0.

EXAMPLE 4