Math 154 Elementary Algebra
Caspers
Name___________________________
Date____________________________
Intercepts and Slope — 4.3
Intercepts
2y   x  4
Below is a graph of
6
5
4
3
State the x-intercept:
2
1
0
-6
-5
-4
-3
-2
-1
-1
0
-2
1
2
3
4
5
6
State the y-intercept:
-3
-4
-5
-6
The x-intercept of an equation is the point where the graph crosses the x-axis. The x-intercept’s y-value is always 0. In other
words, the x-intercept is always a point of the form ( ? , 0 ). To find the x-intercept from an equation, plug in 0 for y and then find
the x-value that goes with y = 0. Remember, for the x-intercept, you are looking for an x-value.
The y-intercept of an equation is the point where the graph crosses the y-axis. The y-intercept’s x-value is always 0. In other
words, the y-intercept is always a point of the form ( 0 , ? ). To find the y-intercept from an equation, plug in 0 for x and then find
the y-value that goes with x = 0. Remember, for the y-intercept, you are looking for a y-value.
Find the x-intercept, the y-intercept, and a different third “check” point, then graph the line.
1.
4x  2 y   1 2
6
Find the x-intercept:
5
4
3
2
Find the y-intercept:
1
0
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
5
6
-2
-3
Check point:
-4
-5
-6
Page 1 of 7
Slope
All linear equations in two variables have a constant rate of change or slope. The slope of a line is a ratio of the vertical change to
the horizontal change between two selected points on a line. In other words, slope is
rise
run
t h e ch an g e i n t h e y-valu es
the change in the x-values
or
2.
or
vertical change
horizontal change
or
the change up or down
.
the change left or right
Find the slope of the line graphed.
8
7
By counting the units of rise—vertical (y) change—and the units
of run—horizontal (x) change—between any two points:
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
3.
Find the slope of the line graphed.
8
7
By counting the units of rise—vertical (y) change—and the units
of run—horizontal (x) change—between any two points:
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
Page 2 of 7
The slope of a line can also be found by using the slope formula. If two points on the line are labeled the first ordered pair:
 x1 , y1  and the second ordered pair:  x2 , y2  , then the slope of the line can be found by the following formula:
y2  y1
x2  x1
It does not matter which ordered pair is labeled the first ordered pair and which is labeled the second ordered pair.
m 
4.
Find the slope of the line graphed.
8
By counting the units of rise—vertical (y) change—and the units
of run—horizontal (x) change—between any two points:
7
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
By using two points on the line and the slope formula:
8
-2
-3
-4
-5
-6
-7
-8
5.
Find the slope of the line graphed.
8
By counting the units of rise—vertical (y) change—and the units
of run—horizontal (x) change—between any two points:
7
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
By using two points on the line and the slope formula:
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
Graphing a linear equation means drawing a line that shows all of the points that satisfy the equation. There is more than
one way to draw this line.
1. One way is to find any two points that work and then another check point.
2. Another way is to specifically find the x-intercept and the y-intercept and then another check point.
3. The third way is to plot one point and then use the slope to find two other points.
Page 3 of 7
For problems 6 through 8, graph the line with the given slope that goes through the given point.
2
6.
Through  0 ,  2 
with m 
3
8
7
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
7.
Through
  3 , 5
with m   2
8
7
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
8.
Through
  3 , 0
with m  
2
5
8
7
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
Page 4 of 7
Slopes of Horizontal and Vertical Lines
Find the slope of the line through the given points.
9.
0 ,  2
and
3 ,  2
A graph of this line:
8
7
6
5
Slope:
4
3
Y-intercept:
2
1
Equation of the line:
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
2
3
4
5
6
7
8
2
3
4
5
6
7
8
-2
-3
-4
-5
-6
-7
-8
The slope of any horizontal line is 0. So, if the equation is y = “a constant”, then the slope, m, is 0.
Find the slope of the line through the given points.
10.
 3 ,  1
and
3 , 2
A graph of this line:
8
7
6
5
Slope:
4
3
X-intercept:
2
1
Equation of the line:
-8
-7
-6
-5
-4
-3
-2
0
-1 -1 0
1
-2
-3
-4
-5
-6
-7
-8
The slope of any vertical line is undefined. So, if the equation is x = “a constant”, then the slope, m, is undefined.
Page 5 of 7
11.
The graphs of two lines are shown below. Find the slope of each line.
8
7
Line 1
6
5
4
3
2
Line 1
Line 2
1
0
-8 -7 -6 -5 -4 -3 -2 -1-1 0
Line 1
1
2
3
4
5
6
7
8
-2
-3
Line 2
-4
-5
-6
Line 2
-7
-8
Parallel Lines
Any non-vertical lines that have the same slope, but different y-intercepts are parallel. Any two distinct vertical lines are parallel to
each other. For example, line 1 that has a slope m1  53 and a y-intercept of  0 , 2  is parallel to line 2 that has a slope of m2  53
and a y-intercept of  0 ,  6  . Slope indicates how a line “leans”, and the y-intercept indicates “where” a line is located.
12.
The graphs of two lines are shown below. Find the slope of each line.
Perpendicular Lines
Two lines whose slopes are opposite reciprocals are perpendicular to each other. For example, line 1 that has a slope m1   13 is
perpendicular to line 2 that has a slope of m2  3 . The y-intercept of these lines is irrelevant to the fact that they are perpendicular.
In the case of a vertical line and a horizontal line: any vertical line is perpendicular to any horizontal line.
Page 6 of 7
For problems 13 through 16, the slopes of two distinct lines are listed. Use the slopes to determine whether each pair of lines is
parallel, perpendicular, or neither parallel or perpendicular.
13.
Line 1: m1  3 ,
Line 2: m2  3
14.
Line 1: m1  2 ,
15.
Line 1: m1  
Line 2: m2   2
1
,
3
Line 2: m2   3
1
5
Use the pair of points given for each line to find the slope of that line and then determine if the pair of lines is parallel,
perpendicular, or neither parallel or perpendicular.
17.
Line 1:   12 , 1 and  34 , 2 
Line 2:  7 , 3 and  4 ,  3
16.
Line 1: m1  5 ,
18.
Line 1:  0 , 6  and 1 , 4 
Line 2: m2  
Line 2:  4 , 5 and  0 , 13
4.1–4.3 Review Problems
19.
Graph  3 y  6
20.
Graph y  
8
8
7
7
6
6
5
5
4
4
3
3
2
2
0
-8 -7 -6 -5 -4 -3 -2 -1-1 0
0
-8 -7 -6 -5 -4 -3 -2 -1-1 0
x
1
1
21.
3
5
1
2
3
4
5
6
7
8
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
-7
-8
-8
Graph the line that goes through
1 ,  2  and
has a slope, m   3
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
8
7
6
5
4
3
2
1
0
-8 -7 -6 -5 -4 -3 -2 -1-1 0
-2
-3
-4
-5
-6
-7
-8
 5x  2 0  2 y
22.
Find the x-intercept and the y-intercept.
23.
Find the slope of the line between two points given.
  4 , 6
and
1 , 5 
Page 7 of 7