Unit 4 Study Guide ANSWERS
1. Identify each graph as being a non-linear function, linear function, or not a function.
Graph A
Graph B
Graph C
y
3
–3
–2
3
5
2
4
1
3
–1
–1
y
y
1
2
2
1
2
3 x
–2
1
–1
–1
1
2
3
4 x
–2
–2
–1
–3
1
–1
x
–3
Graph A: non-linear function
Graph B: linear function
Graph C: not a function
In a function, each domain value is paired with exactly one range value.
Graph A is a function, but it is not linear.
Graph B is a function and a line.
Graph C is not a function because each domain value (–0.8) pairs with infinite range values.
2. Tell whether the set of ordered pairs
satisfies a linear function. Explain.
Yes; there is a constant change in x that corresponds to a constant change in y.
In a linear function, a constant change in x means a constant change in y.
+2
+2
+2
x
2
4
6
8
y
3
9
15
21
3. Tell whether the function
+6
+6
+6
constant change in y
is linear. If so, graph the function.
The equation can be written in standard form, so the function is linear.
y  2x  4
y  2x  4
-2x -2x
y  2 x  4
2 x  y  4
Write the equation in standard form. Try to get both variables on the same side.
Subtract 2x from both sides.
y  2 x  4 is equivalent to 2 x  y  4 .
The equation is in the form Ax + By = C (A = -2, B = 1, C = -4).
To graph the function, choose three values of x and use them to generate ordered pairs.
x
y = 2x –4
(x, y)
Plot the points and connect them with a straight line.
y
0 y = 2(0) – 4 = -4
(0, –4)
8
1 y = 2(1) – 4 = -2
(0, –2)
6
2 y = 2(2) – 4 = 0
(0, 0)
4
2
(2, 0)
–8
–6
–4
–2
–2
–4
–6
–8
(1,
2 –2)
4
(0, –4)
6
8 x
4.
Find the x- and y-intercepts.
5.
Find the x- and y-intercepts of
.
x-intercept: –5, y-intercept: 5
y
8
6
To find the x-intercept, replace y with 0 and
solve for x; to find the y-intercept, replace x with
0 and solve for y.
x-intercept:
y-intercept:
-4x + 4(0) = 20
-4(0) + 4y = 20
-4x = 20
4y = 20
20
20
x
y
4
4
x = -5
y=5
4
2
–8
–6
–4
–2
–2
2
4
6
x
8
–4
–6
–8
x-intercept: –1, y-intercept: –3
The graph intersects the x-axis at (–1, 0).
The graph intersects the x-axis at (–1, 0).
The x-intercept is –1.
The graph intersects the y-axis at (0, –3).
The y-intercept is –3.
6. Find the slope of the line.
y
7. Find the slope of the line.
(7, 7)
(–8, 8) (–4, 8)
8.
Tell whether the slope of the line is
positive, negative, zero, or undefined.
y
7
y
8
6
6
6
4
(3, 4)
5
4
4
2
2
–8
–6
–4
–2
2
–2
4
6
8
x
3
2
–8
–6
–4
–6
–8
–4
–2
–2
2
4
6
8
x
–4
–6
–8
3
0
4
To find the slope, use the coordinates of two points on the line.
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5 x
–2
–3
undefined
Starting at one point, count the units down (negative units) or up (positive units) and to the right (positive
units) or to the left (negative units) to arrive at the other point. The units up or down are the rise. The
units to the right or to the left are the run.
Write a fraction with the rise in the numerator and the run in the denominator. Simplify the fraction.
y2  y1
x2  x1
y  y1
88
0
m 2


. The slope is 0.
x2  x1 8  4 4
m
A line has positive slope if it rises from left to right.
A line has negative slope if it falls from left to right.
A line has zero slope if it is a horizontal line.
A line has undefined slope if it is a vertical line.
9. Find the slope of the line that contains
m
13
11
.
Find the slope of the line described by x + 4y = –8.
10.

y2  y1
x2  x1
6  (7 )
m
2  (9)
Use the slope formula.
m
m
and
Substitute (-9, -7) for
(2, 6) for
13
11
1
4
Rewrite the equations in slope-intercept form.
( x1 , y1 ) and
( x2 , y2 ) .
x + 4y = -8
-x
-x
4y = -x – 8
4 4 4
Simplify.
y
1
x 2
4
The slope is the fraction in front of x.
11. Graph the line with the slope  and y-intercept 2.
1
3
y
5
1
Plot the y-intercept 2 on the graph at (0, 2). The slope is  , so from the y3
intercept, rise –1 units and run 3 units. Plot another point. Connect the
points to graph the line.
4
3
y-intercept
run
2
rise
1
–5
–4
–3
–2
–1
–1
1
2
3
4
x
5
–2
–3
–4
–5
12. Write the equation that describes the line with slope = 1 and y-intercept = 2 in slope-intercept form.
y=x+2
The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. Substituting 1 for the slope and
2 for the y-intercept gives y = x + 2.
13. Write the equation that describes the line in slope-intercept form. slope = 4, point (3, –2) is on the line
y = 4x – 14
If you are given the slope and one point, you can write an equation by starting with point-slope form
y  y1  m( x  x1 ) then rewriting the equation in y = mx + b.
y  y1  m( x  x1 )
y – (-2) = 4(x – 3)
y + 2 = 4x – 12
-2
-2
y = 4x – 14
So, the equation of the line in slope-intercept form is y = 4x – 14.
14. Write the equation
in slope-intercept form. Then graph the line described by the equation.
4
5
y=  x–4
y
4x + 5y = -20
-4x
-4x
5y = -4x – 20
5 5 5
10
8
6
4
2
–8
–6
–4
–2
–2
2
4
6
8
x
y= 
–4
–6
–8
–10
m= 
4
x–4
5
4
, b = -4
5
Plot (0, -4). Count 4 down and 5 right, and plot
another point. Draw a line connecting the two points.
15. The river level of a river is 25 feet and it is receding at a rate of 0.25 foot per day. Write an equation that represents the
water level, w, after d days. Identify the slope and y-intercept and describe their meanings. In how many days will the
water level be 20 feet?
w = -0.25d + 25
The slope is -0.25, and this is the rate at which the water level is receding.
The y-intercept is 25, and this is the water level after 0 days.
In 20 days, the water level will be 20 feet.
Step 1: Write an equation that represents the water level, w, after d days.
Water level
is
receding at 0.25 foot
per day
starting at
w
=
-0.25
d
+
25 feet
25
An equation is w = -0.25d + 25.
Step 2: Identify the slope and y-intercept and describe their meanings.
The slope is -0.25. This is the rate at which the water level is receding: 0.25 foot per day. The y-intercept is 25. This
is the water level after 0 days, or the starting water level of 25 feet.
Step 3:
w = -0.25d + 25
20 = -0.25d + 25
Substitute 20 for w in the equation.
d = 20
Solve for d.
In 20 days, the water level will be 20 feet.
16. Graph the line with a slope of  3 that contains the point (–3, –1).
2
y
To graph the line, plot the given point (–3, –1). Then, use the slope 
10
8
additional points. Use a rise of –2 (moving up if the slope is positive and down if the
slope is negative) and a run of 3 (moving right if positive and left if negative) to find
additional points.
6
4
2
–10 –8
–6
–4
–2
–2
2
to plot
3
2
4
6
8
x
–4
–6
–8
–10
17. Write an equation in point-slope form for the line that has a slope of
y–5=
4
7
and contains the point (2, 5).
4
(x – 2)
7
Substitute the point and slope into the point-slope form y  y1  m( x  x1 ) , where m represents the slope and
( x1 , y1 ) represents a point on the line.
18. A linear function has the same y-intercept as
slope of the linear function.
y–intercept = (0, 3), slope =
Step 1 Find the y-intercept.
x  5 y  15
5 y   x  15
1
y   x 3
5
y–intercept = (0, 3)
and its graph contains the point
1
2
Subtract x from both sides.
Divide both sides by 5.
. Find the y-intercept and
Step 2 Find the slope.
y  y1
m 2
x2  x1
5 3 2 1
 
40 4 2
Use the slope formula.
Substitute the given point (4, 5) and the y–intercept (0, 3)
and simplify.
19. The equations of four lines are given. Identify which lines are parallel.
Line 1: y = 5x – 7
1
Line 2: x – 6 y = –1
Line 3: y = 6x – 3
1
Line 4: y – 4 =  5 (x – 9)
Lines 2 and 3 are parallel.
Write all the equations in slope-intercept form (y = mx + b). The equations that have the same
slope but different y-intercepts are parallel lines.
20. Identify the lines that are perpendicular:
Line 1:
Line 2:
Line 3:
Line 4:
y = 6 and x = 1 are perpendicular; y = -⅓x + 3 and y + 2 = 3(x – 1) are perpendicular.
y = 6 is horizontal, and x = 1 is vertical. These lines are perpendicular.
Check if the product of the slopes of the other two lines is -1.
y = -⅓x + 3 has slope = -⅓. y + 2 = 3(x – 1) has slope = 3.
3(-⅓) = -1
The product of the slopes is -1, so these two lines are also perpendicular.
21. Write an equation in slope-intercept form for the line parallel to y = -2x – 2 that passes through the point (9, –2).
y = -2x + 16
Parallel lines have the same slope.
Since the given line has a slope of -2, the parallel line has a slope of -2.
y – y1 = m(x – x1)
Use the point-slope form to write an equation.
y – (–2) = -2(x – 9)
Substitute -2 for m and (9, –2) for (x1, y1).
y + 2 = -2x +18
Distribute -2 on the right side.
y = -2x + 16
Add -2 to both sides.
Monthly Salary ($)
22. Kam is a car saleswoman. The table shows the salary that Kam earns for the number of cars she sells. Use the data to
make a graph. Then, find the slope of the line and explain what it shows.
Monthly Salary
Number of Cars
5000
Earned
Sold
4000
$1400
0
$1700
1
3000
$2000
2
$2300
3
2000
4
$2600
1000
The slope of the line is 300. This means that for every car Kam
sells, she earns $300. Graph the data.
Find the slope of the line using the slope formula
y2  y1
.
x2  x1
The slope of the line is 300. This means that for every car Kam sells, she earns $300.
2
4
6
8
Number of Cars S old
23. Write the equation –4x + y = –21 in slope-intercept form, and then find the slope and y-intercept.
slope-intercept form: y = 4x – 21; slope: 4 ; y-intercept: -21
Rewrite the equation in the form y  mx  b to find the slope and the y-intercept.
–4x + y = –21
+4x
+4x
y = 4x – 21
24. Write the point-slope form of the equation of the line with slope
y+9=
5
(x + 5)
7
5
7
that passes through the point (–5, –9).
Put the point and slope into the point-slope form y  y1  m( x  x1 ) , where m represents the slope
and ( x1 , y1 ) represents a point on the line.
25. Tell whether (6, 7) is a solution of
.
26.
Tell whether (7, 8) is a solution of
No, (6, 7) is not a solution of y  4 x  6 .
Yes, (7, 8) is a solution of y  3 x  7 .
Substitute (6, 7) for (x, y) in y  4 x  6 .
Substitute (7, 8) for (x, y) in y  3 x  7 .
y  4x  6
7  4(6)  6
7  30 , false
y  3x  7
8  3(7)  7
8  14 , true
(6, 7) is NOT a solution of y  4 x  6 .
27. Graph the solutions of the linear inequality
y
10
8
6
4
2
–10 –8
–6
–4
–2
–2
2
4
6
8
10
x
.
(7, 8) is a solution of y  3 x  7 .
.
Step 1. Solve the inequality 4 x  2 y  6 for y.
4 x  2 y  6
+4x
+4x
2y > 4x + 6
2 2 2
y  2x  3
Step 2. Graph the boundary line y  2 x  3 . Use a dashed line for >.
–4
–6
–8
–10
y
Step 3. The inequality is >, so shade above the line.
28. Write an inequality to represent the graph.
y
y  –4x – 3
5
4
3
2
1
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
1
2
3
4
5
x
Use the graph to determine the slope and y-intercept, and then write an
equation in the form y = mx + b. A graph shaded above the line means
x
greater than
and the graph shaded below the line means less than. Use 
or  if the line is solid; use < or > if the line is dashed.
29. A movie theater charges $9.00 for an adult ticket to an evening showing of a popular movie. To help the local animal
shelter, the theater management has agreed to reduce the price of each adult ticket by $0.25 for every can of pet food a
customer contributes to a collection barrel in the theater lobby. Write both an equation in which y represents the cost of an
adult ticket in dollars for a customer who contributes x cans of pet food, and graph of the cost if a customer brings in 2, 5,
8, or 10 cans of pet food?
y = 9 – 0.25x
30. Harry is considering buying a multi-disk compact disk player for $375. He would also like to buy some new compact
disks. They are on sale for $9.99 each. Describe the most appropriate graph (scatter plot or linear equation) to represent
the total cost, before sales tax, for Harry to buy the player and some disks?
The graph should be a scatter plot with the first point at (0, 375) and have a vertical line of
best fit with a slope of 9.99.