This document provides a study guide for an 8th grade algebra test covering linear equations and relationships. It lists 17 concepts to be familiar with, including finding the slope and y-intercept of a line from its graph or equation, graphing linear equations, finding x- and y-intercepts, creating linear equations from graphs or two points, identifying linear and inverse relationships from graphs and tables, identifying parallel and perpendicular lines from graphs and equations, and finding equations of lines parallel or perpendicular to given lines. For each concept, it provides a brief explanation and example.

1. 8th
Grade
Unit
1
Study
Guide
Algebra
What you should know for this test:
Check
Goal
Concept
1. I
can
find
the
slope
Find the y-intercept by seeing where the line crosses the y-
axis.
and
y-­‐intercept
of
an
equation
from
its
Find the slope by using two points on the line and counting
graph.
the rise over the run.
2. I
can
find
the
slope
When an equation is in y = mx + b format, the ‘m’ is the
and
y-­‐intercept
of
slope, and the ‘b’ is the y-intercept.
an
equation.
In y = 2/3x + 4, 2/3 is the slope and 4 is the y-intercept.
3. I
can
graph
a
linear
Take the ‘b’ in your equation and find that number on the y-
equation
in
axis. Place your first point there. Then use the slope as your
rise over run. Go up and over to place your next point.
y
=
mx+b
format.
In y = 2/3x + 4, start at 4 on the y-axis. Then go up 2, over 3
and put your next point.
You only need to do this once to see how your line needs to
be drawn.
4. I
can
find
the
x
and
Take an equation, for example, y = 5x + 10.
y
intercepts
of
a
To find the y-intercept, substitute 0 for x: y = 5(0) + 10, so y
linear
equation.
= 10. The y-intercept is 10. (Or you may have noticed that 10
is in the ‘b’ spot, so it has to be the y-intercept.)
To find the x-intercept, substitute 0 for y: 0 = 5x+10, so x = -
2. The x-intercept is -2.
5. I
can
create
a
linear
Find the y-intercept and the slope of the line, then plug them
equation
from
a
in for m and b in y = mx + b.
graph.
6. I
can
create
a
linear
If you know two points, you can find the slope by finding the
equation
from
two
difference in the y’s over the difference in the x’s. Then plug
in the slope you found, along with one of the x and y pairs,
points
on
the
line
of
into y = mx + b.
its
graph.
Example, (2,3) and (4,5). The difference between 3 and 5 is
2, and the difference between 2 and 4 is 2, so the slope is 2/2
or 1.

2. 8th
Grade
Unit
1
Study
Guide
Algebra
Plug in 1 as the slope, 2 as x, and 3 as y into y =mx + b:
3 = 1(2) + b. Substitute numbers for x, y, and m
3=2+b Simplify 1 x 2
1=b Subtract 2 from both sides; your equation
is y = 1x + 1
7. I
can
create
a
linear
This is just like finding the equation from two points, except
equation
from
1
you get to skip the step where you find the slope because
you’re already given it.
point
and
the
slope
of
the
equation.
For example: (2,5) with a slope of 3
Start with y = mx + b and plug in what you know: your slope
is 3, your x is 2, and your y is 5.
5 = 3(2) + b
5=6+b
-1 = b
So your equation is y = 3x -1
8. I
can
recognize
a
The graphs of a linear relationship are always a straight line.
linear
relationship
from
a
graph.
9. I
can
recognize
a
Linear relationships can be easily discovered in tables. If all
linear
relationship
the x values are going up at the same rate as each other, and
all the y values are going up or down at the same rate as each
from
a
table
of
other, then it is a linear relationship.
values.
10. I
can
recognize
an
The graphs of an inverse relationship are always curving
inverse
relationship
downward. They drop a lot at first, then even out toward the
x-axis.
from
a
graph.

3. 8th
Grade
Unit
1
Study
Guide
Algebra
11. I
can
recognize
an
Inverse relationships can be easily discovered in tables. As x
inverse
relationship
values go up, y values go down, at first by a lot, and then
much more slowly. Another way to recognize inverse
from
a
table
of
relationships is to multiply all the x and y pairs; if they all
values.
multiply to be the same number then it is an inverse
relationship.
12. I
can
identify
Parallel lines on a graph are easy to spot. They are just
parallel
lines
from
a
straight lines that never cross. Their slope is always the same
(which is why they never cross).
graph.
13. I
can
identify
Parallel lines have the same slope:
parallel
lines
from
y = 4x + 3 and y = 4x – 4 are parallel—same slope
y = 7x – 9 and y = 7x + 9 are parallel—same slope
their
equations.
y = 1/2x – 3 and y = 2x – 4 are not parallel—different slope
14. I
can
identify
Perpendicular lines on a graph are easy to spot. They are just
perpendicular
lines
straight lines that cross at exactly a 90 degree angle (they
make a big plus sign). Their slopes are negative reciprocals
from
a
graph.
of each other (the fraction flips and you change the sign).
y = 2x and y = -1/2x are perpendicular
Other perpendicular lines:
y = 4x + 2 and y = -1/4x + 4
y = 1/3x + 6 and y = -3x + 2
y = -1/5x + 3 and y = 5x - 2
15. I
can
identify
You can identify perpendicular lines by their slopes. If the
perpendicular
lines
slopes are negative reciprocals, then the two lines are
perpendicular.
from
their
equations.
y = -4x + 2 and y = 1/4x + 7
y = 9x – 4 and y = -1/9 – 3

4. 8th
Grade
Unit
1
Study
Guide
Algebra
16. I
can
find
the
This is just like when you create an equation using 1 point
equation
of
a
line
and the slope. Since you know the equation of a line that is
parallel, you just take that slope and plug it into your new
parallel
to
another
equation. Then take the point you are given and plug in the x
line
from
just
1
and y values.
point
and
the
equation
of
the
Find the equation of a line that is parallel to y = 3x + 5 and
parallel
line.
goes through the point (4, 2)
Since the equations are parallel, the slope of your new
equation must be 3. The x value is 4 and the y value is 2.
Plug these values into y = mx + b.
2 = 3(4) + b
2 = 12 + b
-10 = b
The new equation is y = 3x – 10
17. I
can
find
the
This is just like when you create an equation using 1 point
equation
of
a
line
and the slope. Since you know the equation of a line that is
perpendicular, you can take that slope, find it’s negative
perpendicular
to
reciprocal, and plug it into your new equation as your new
another
line
from
slope. Then take the point you are given and plug in the x and
just
1
point
and
the
y values.
slope.
Find the equation of a line that is perpendicular to y = 5x + 1
and goes through the point (10, 2)
Since the equations are perpendicular, the slope of your new
equation must the negative reciprocal of 5, which would be -
1/5. The x value is 10 and the y value is 2.
Plug these values into y = mx + b.
2 = -1/5(10) + b
2 = -2 + b
4=b
The new equation is y = -1/5x + 4