Creating a Table of Values and
Finding the Intercepts
§  There are 3 ways to graph a linear
relation:
ú  From the definition of y = mx + b.
ú  By creating a table of values.
ú  Plotting the x- and y-intercepts.
§  We have already graphed lines from the
equation y = mx + b, using the yintercept and the slope.
§  Now we will attempt the other two
methods.
§  To create a table of values:
ú  Choose at least two values for x. (If you
ú 
ú 
ú 
ú 
choose 5 values; two positive values, zero,
and two negative values, you will get a
better picture.)
Substitute these values into the equation to
find the matching y value.
Plot this ordered pair.
Continue for any remaining x-values.
Join the points with a line. (Use a ruler!)
§  Graph the line y = -3x + 1 by completing
the table of values.
x
-2
-1
0
1
2
y
y
10
9
8
7
6
5
4
3
2
1
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-2
-3
-4
-5
-6
-7
-8
-9
-10
1
x
2 3 4 5 6 7 8 9 10
§  Graph the line y = ½x + 5 by completing
the table of values.
x
y
y
10
9
8
7
6
5
4
3
2
1
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-2
-3
-4
-5
-6
-7
-8
-9
-10
1
x
2 3 4 5 6 7 8 9 10
§  This method would be best for standard form.
§  It allows you to find the two points where the
line crosses the x- and y-axes.
§  To find the x-intercept:
ú  Let y = 0.
ú  Solve for x.
ú  Plot this point.
§  To find the y-intercept:
ú  Let x = 0.
ú  Solve for y.
ú  Plot this point.
§  Join the points with a line.
§  Graph the line 2x – 5y – 10 = 0 by
finding the intercepts.
To find x-intercept:
Let y = 0.
2x – 5(0) – 10 = 0
2x – 10 = 0
2x = 10
x=5
(5, 0)
To find y-intercept:
Let x = 0.
2(0) – 5y – 10 = 0
-5y – 10 = 0
-5y = 10
y = -2
(0, -2)
y
10
9
8
7
6
5
4
3
2
1
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-2
-3
-4
-5
-6
-7
-8
-9
-10
1
x
2 3 4 5 6 7 8 9 10
§  Graph the line 3x + 6y + 12 = 0 by
finding the intercepts.
10
9
8
7
6
5
4
3
2
1
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-2
-3
-4
-5
-6
-7
-8
-9
-10
y
1
x
2 3 4 5 6 7 8 9 10
§  Textbook
§  Worksheet