Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
90
Graphing Using Intercepts
The easiest way to plot two points on a line quickly
5.18 What are the x- and y-intercepts of a graph?
An x-intercept is a value at which a graph intersects the x-axis. It is usually
reported as a signed value, indicating the location of the intersection point as
though the x-axis were a number line. Similarly, the y-intercept reports the
y-value at which a graph intersects the y-axis.
5.19 If a graph has x-intercept –1 and y-intercept 9, what are the coordinates of
those intercepts?
All x-intercepts are located on the x-axis, which has equation y = 0 (according
to Problem 5.4). Therefore, the coordinate of an x-intercept must always have
a y-value of 0. Similarly, all y-intercepts are located on the y-axis, which has the
equation x = 0. Therefore, the coordinate of a y-intercept must always have an
x-value of 0. The graph in this problem must pass through the points (–1,0) and
(0,9).
Note: Problems 5.20–5.21 refer to the equation x – 3y = 12.
5.20 Identify the x- and y-intercepts of the linear graph.
According to Problem 5.19, an x-intercept has a corresponding y-value of 0.
Substitute y = 0 into the equation and solve for x to identify the x-intercept.
The point at which a graph intersects the y-axis has an x-value of 0. Substitute
x = 0 into the equation to determine the y-intercept.
The equation x – 3y = 12 has x-intercept 12 and y-intercept –4.
Note: Problems 5.20–5.21 refer to the equation x – 3y = 12.
5.21 Graph the equation using the intercepts calculated in Problem 5.20.
According to Problem 5.20, the intercepts are x = 12 and y = –4, so the graph
passes through the points (12,0) and (0,–4). Plot those points on the coordinate
plane and connect them to graph the line, as illustrated by Figure 5-13.
In other
words, if a graph
has an x-intercept
of –4, that means it
passes through the x-
axis four units left
of the origin.
All lines
(excluding
horizontal and
vertical lines)
have exactly one
x-intercept and one
y-intercept, so after
you’ve found one x-
intercept for this line,
you’ve found all of
them.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
91
Figure 5-13: The graph of x – 3y = 12 has x-intercept 12 and y-intercept –4.
Note: Problems 5.22–5.23 refer to the equation 2x – y = 0.
5.22 Identify the x- and y-intercepts of the linear graph.
Use the technique described in Problem 5.20; substitute y = 0 into the equation
to calculate the x-intercept and substitute x = 0 to calculate the y-intercept.
The x- and y-intercepts of the graph are both 0.
Note: Problems 5.22–5.23 refer to the equation 2x – y = 0.
5.23 Explain why you cannot graph the equation using intercepts alone.
Graph the equation using a different technique.
According to Problem 5.22, the x- and y-intercepts are both 0, which means the
graph passes through the origin. Unfortunately, when a linear graph passes
through the origin, its x- and y-intercepts are represented by the same point:
(0,0). However, two distinct points are required to graph a line.
To overcome this obstacle, substitute a nonzero x- or y-value into the equation
and solve for the remaining variable to identify a second point on the line. For
instance, substitute x = 2 into the equation and solve for y.
The prob-
lem isn’t that
the intercepts
are both equal—
that’s normally ne.
The problem is that
they’re both 0. If the
intercepts were both
3, for example, then
you’d know the graph
passed through the
points (3,0) and
(0,3). Those are
two different
points.
Chapter Five — Graphing Linear Equations in Two Variables
The Humongous Book of Algebra Problems
92
Because y = 4 when x = 2, the point (2,4) also belongs to the graph. Connect this
point and (0,0) to graph the line, as illustrated by Figure 5-14.
Figure 5-14: The graph of 2x – y = 0 passes through the points (0,0) and (2,4).
Note: Problems 5.24–5.25 refer to the equation 5x = 3y + 16.
5.24 Identify the x- and y-intercepts of the linear graph.
Use the technique described in Problem 5.20; substitute y = 0 into the equation
to calculate the x-intercept and substitute x = 0 to calculate the y-intercept.
The equation has x-intercept and y-intercept .
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