- Using the Slope and Intercept to Graph a Line
When a linear equation is given in the general form, ax+by=c ax+by=c , we observed that an efficient graphical approach was the intercept method. We let x=0 x=0 and computed the corresponding value of y y , then let y=0 y=0 and computed the corresponding value of x x .
When an equation is written in the slope-intercept form, y=mx+b y=mx+b , there are also efficient ways of constructing the graph. One way, but less efficient, is to choose two or three x-values x-values and compute to find the corresponding y-values y-values . However, computations are tedious, time consuming, and can lead to errors. Another way, the method listed below, makes use of the slope and the y-intercept y-intercept for graphing the line. It is quick, simple, and involves no computations.
- Plot the y-intercept y-intercept (0, b) (0, b) .
- Determine another point by using the slope m slope m .
- Draw a line through the two points.
Recall that we defined the slope m slope m as the ratio y 2 − y 1 x 2 − x 1 y 2 − y 1 x 2 − x 1 . The numerator y 2 − y 1 y 2 − y 1 represents the number of units that y y changes and the denominator x2−x1 x 2 x 1 represents the number of units that x x changes. Suppose m= p q m= p q . Then p p is the number of units that y y changes and q q is the number of units that x x changes. Since these changes occur simultaneously, start with your pencil at the y-intercept y-intercept , move p p units in the appropriate vertical direction, and then move q q units in the appropriate horizontal direction. Mark a point at this location.
Graph the following lines.
y= 3 4 x+2 y= 3 4 x+2
- Step 1: The y-intercept y-intercept is the point ( 0, 2 ) ( 0, 2 ) . Thus the line crosses the y-axis y-axis 2 units above the origin. Mark a point at ( 0, 2 ) ( 0, 2 ) .
- Step 2: The slope, m m , is 3 4 3 4 . This means that if we start at any point on the line and move our pencil 3 3 units up and then 4 4 units to the right, we’ll be back on the line. Start at a known point, the y-intercept ( 0, 2 ) y-intercept ( 0, 2 ) . Move up 3 3 units, then move 4 4 units to the right. Mark a point at this location. (Note also that 3 4 = −3 −4 3 4 = −3 −4 . This means that if we start at any point on the line and move our pencil 3 3 units down and 4 4 units to the left, we’ll be back on the line. Note also that 3 4 = 3 4 1 3 4 = 3 4 1 . This means that if we start at any point on the line and move to the right 1 1 unit, we’ll have to move up 3/4 3/4 unit to get back on the line.)
- Step 3: Draw a line through both points.

y=− 1 2 x+ 7 2 y=− 1 2 x+ 7 2
- Step 1: The y-intercept y-intercept is the point ( 0, 7 2 ) ( 0, 7 2 ) . Thus the line crosses the y-axis y-axis 7 2 units 7 2 units above the origin. Mark a point at ( 0, 7 2 ) ( 0, 7 2 ) , or ( 0, 3 1 2 ) ( 0, 3 1 2 ) .
- Step 2: The slope, m m , is − 1 2 − 1 2 . We can write − 1 2 − 1 2 as −1 2 −1 2 . Thus, we start at a known point, the y-intercept y-intercept ( 0, 3 1 2 ) ( 0, 3 1 2 ) , move down one unit (because of the −1 −1 ), then move right 2 2 units. Mark a point at this location.
- Step 3: Draw a line through both points.

y= 2 5 x y= 2 5 x
- Step 1: We can put this equation into explicit slope-intercept by writing it as y= 2 5 x+0 y= 2 5 x+0 .
The y-intercept y-intercept is the point ( 0, 0 ) ( 0, 0 ) , the origin. This line goes right through the origin.
- Step 2: The slope, m m , is 2 5 2 5 . Starting at the origin, we move up 2 2 units, then move to the right 5 5 units. Mark a point at this location.
- Step 3: Draw a line through the two points.
y=2x−4 y=2x−4
- Step 1: The y-intercept y-intercept is the point ( 0, −4 ) ( 0, −4 ) . Thus the line crosses the y-axis 4 y-axis 4 units below the origin. Mark a point at ( 0, −4 ) ( 0, −4 ) .
- Step 2: The slope, m m , is 2. If we write the slope as a fraction, 2= 2 1 2= 2 1 , we can read how to make the changes. Start at the known point ( 0, −4 ) ( 0, −4 ) , move up 2 2 units, then move right 1 1 unit. Mark a point at this location.
- Step 3: Draw a line through the two points.
Use the y-intercept y-intercept and the slope to graph each line.
y= −2 3 x+4 y= −2 3 x+4

y= 3 4 x y= 3 4 x

For the following problems, graph the equations.
y= 2 3 x+1 y= 2 3 x+1

y= 1 4 x−2 y= 1 4 x−2

y=5x−4 y=5x−4

y=− 6 5 x−3 y=− 6 5 x−3

y= 3 2 x−5 y= 3 2 x−5

y= 1 5 x+2 y= 1 5 x+2

y=− 8 3 x+4 y=− 8 3 x+4

y=− 10 3 x+6 y=− 10 3 x+6

y=1x−4 y=1x−4

y=−2x+1 y=−2x+1

y=x+2 y=x+2

y= 3 5 x y= 3 5 x

y=− 4 3 x y=− 4 3 x

y=x y=x

y=−x y=−x

3y−2x=−3 3y−2x=−3

6x+10y=30 6x+10y=30

x+y=0 x+y=0

((Reference)) Determine the slope and
y-intercept
y-intercept
of the line
−4y−3x=16
−4y−3x=16
.
((Reference)) Find the slope of the line passing through the points
(−1, 5)
(−1, 5)
and
(2, 3)
(2, 3)
.
"Elementary Algebra covers traditional topics studied in a modern elementary algebra course. Written by Denny Burzynski and Wade Ellis, it is intended for both first-time students and those […]"