4.4 The Slope-Intercept Form of
the Equation of a Line
Slope-Intercept Form
The slope-int. form of an eqn. of a line with slope
m and y-intercept b is
y = mx + b
slope y-int
Ex. Find the eqn. of a line with slope 4 and y-int 1.
y = mx + b
m = 4, b = 1
y = 4x + 1
Ex. Find the slope and y-intercept of y = ¼x – 2.
y = ¼x – 2
y = ¼x + (-2)
m = ¼, b = -2 or (0, -2)
Ex. Find the slope and y-intercept of y = -5x.
y = -5x + 0
m = -5, b = 0 or (0, 0)
Ex. Find the slope and y-intercept of 4x – 2y = 8.
We need to solve for y first (put in slope-int. form y=mx + b)
4x – 2y = 8
4x – 2y – 4x = 8 – 4x
-2y = -4x + 8
-2y = -4x + 8
-2 -2 -2
y = 2x – 4
m = 2, b = -4
Graphing Using Slope and Y-Intercept
y = mx + b
slope y-int: b or (0, b)
1)
2)
3)
4)
Solve the eqn. for y (put in slope-int form)
Plot the y-int. b or (0, b)
rise
Use slope to plot next point m 
run
Draw line
Ex. Graph y = 2x – 1 using slope and y –intercept.
y
3
2
run= 1
1
-3
rise-1= 2
-2
1
-1
-2
-3
2
3
x
1) y = 2x – 1
m = 2, b = -1 or (0, -1)
2) Plot (0, -1)
3) Using rise = 2 and run = 1,
go up 2 units and right 1
unit to end up at (1, 1) for
a 2nd point.
rise
2 (up 2)
m

run
1 (right 1)
4) Draw line
Ex. Graph y = -½x using slope and y –intercept.
y
3
2
1
-3
-2
-1
rise = -1
1
-1
-2
-3
run = 2
2
3
x
1) y = -½ x + 0
m = -½, b = 0 or (0, 0)
2) Plot (0, 0)
3) Using rise = -1 and run = 2,
go down 1 unit and right 2
units to end up at (2, -1)
for a 2nd point.
rise  1 (down 1)
m

run
2 (right 2)
4) Draw line
Ex. Graph 5x – 3y = 9 using slope and y –intercept.
y
3
run = 3
2
1
rise = 5
-3
-2
-1
1
-1
-2
-3
2
3
x
1) Solve for y:
5x – 3y = 9
5x – 3y – 5x = 9 – 5x
-3y = -5x + 9
-3y = -5x + 9
-3 -3 -3
y = 5/3 x – 3
m = 5/3, b = -3 or (0, -3)
2) Plot (0, -3)
3) Using rise = 5 and run = 3, go
up 5 units and right 3 units to
end up at (3, 2) for a 2nd point.
m
rise
5 (up 5)

run
3 (right 3)
4) Draw line
Ex. Graph the following lines. Are the lines
parallel, perpendicular, or neither? Explain why.
y
y = 3x + 1
y = 3x – 3
3
2
1
-3
-2
-1
1
-1
-2
2
3
x
1) y = 3x + 1
m = 3, b = 1
2) y = 3x – 3
m = 3, b = -3
-3
The lines are parallel
because they have the
same slope, but
different y-intercepts.
Ex. Write an eqn. in slope-int form (y=mx+b) of the
line described:
The y-intercept is -7 and the line is perpendicular to the line
-¼x + y = 5.
1) Find slope by putting -¼x + y = 5 in slope-int. form.
-¼x + y = 5
-¼x + y + ¼x = 5 + ¼x
y = ¼x + 5
Since m = ¼, the slope of the perp. line is m = -4 (neg. recip.)
2) Now, b = -7 (given to us) and m = -4 (from step 1) so our line
is:
y = mx + b
y = -4x – 7