Section 4.1 solutions
1) f(x) = 3x – 6
a) slope – The slope is m = 3 (the number in front of the x)
b) y-intercept - The y-intercept = (0,-6)
c) x-intercept (if any)
set function equal to 0 and solve for x
0 = 3x – 6
6 = 3x
2=x
The x-intercept is the point (2,0)
d) sketch a graph
e) Determine whether the function is always increasing, always decreasing, or constant
This graph is always increasing. It never is flat (constant), nor does it ever go down as I move from left
to right.
3) g(x) = -2x – 8
a) slope - the slope is m = -2 (the number in front of the x)
b) y-intercept - the y-intercept is (0,-8)
c) x-intercept (if any)
0 = -2x – 8
8 = -2x
-4 = x
The x-intercept is the point (-4,0)
d) sketch a graph
e) Determine whether the function is always increasing, always decreasing, or constant
The graph Is always decreasing, it never is flat (constant), nor does it ever go up as I move from left to
right.
5) f(x) = 7
You may rewrite this as f(x) = 0x + 7
a)
b)
c)
d)
if it helps
slope The slope is m = 0 (the number in front of the x that I added)
y-intercept - the y-intercept is (0,7)
x-intercept (if any) The graph is parallel to the x-axis and there is no x-intercept.
sketch a graph
e) Determine whether the function is always increasing, always decreasing, or constant
The graph is flat, we say the graph of the function is constant.
2
3
7) 𝑔(𝑥) = 𝑥 − 4
a) slope – the slope is
2
3
b) y-intercept - the y-intercept is (0,-4)
c) x-intercept (if any)
2
0 = 3 𝑥 − 4 now multiply by 3 to clear fraction
d) sketch a graph
2
3*0 = 3 ∗ 3 𝑥 − 3 ∗ 4
0 = 2x – 12
12 = 2x
6=x
The x-intercept (6,0)
e) Determine whether the function is always increasing, always decreasing, or constant
The graph of the function is always increasing.
9) 𝑓(𝑥) =
−𝑥
3
+6
1
3
You may rewrite this as 𝑓(𝑥) = − 𝑥 + 6 if it helps
1
a) slope – the slope is m = − 3
b) y-intercept - the y-intercept is (0,6)
c) x-intercept (if any)
−1
0 = 3 𝑥+6 multiply by (-3) to clear fraction
−3 ∗ 0 = −3 ∗
−1
𝑥
3
d) sketch a graph
+ −3 ∗ 6
0 = x – 18
18 = x
The x-intercept (18,0)
e) Determine whether the function is always increasing, always decreasing, or constant
The graph of the function is always decreasing.
11) Suppose f(x) = 3x – 6 and g(x) = -2x + 4
a) Solve f(x) = 0
3x – 6 = 0
3x = 6
Solution x = 2
c) Solve f(x) = g(x)
3x – 6 = -2x + 4
+2x
+2x
5x – 6 = 4
+6 +6
5x = 10
Solution x = 2
b) Solve f(x) > 0
3x – 6 > 0
3x > 6
Solution x > 2
d) Solve f(x) < g(x)
3x – 6 < -2x + 4
+2x
+2x
5x – 6 < 4
+6 +6
5x < 10
Solution x < 2
e) graph f(x) and g(x) on the same graph, and label the solution to the equation f(x) = g(x)
For the f(x) function plot -6 on the y-axis then go up 3 right 1
For the g(x) function plot 4 on the y-axis and go down 2 right 1
The x-coordinate of the point of intersection is x = 2 (answer from part c)
Get the y-coordinate of the point of intersection by plugging x = 2 into either function
f(2) = 3(2) – 6
f(2) = 0
13) Suppose f(x) = x-3 and g(x) = 2x + 4
a) Solve f(x) = 0
x- 3 = 0
x=3
c) Solve f(x) = g(x)
x – 3 = 2x + 4
-2x - 2x
-1x – 3 = 4
+3
+3
-1x = 7
b) Solve f(x) > 0
x–3>0
x>3
d) Solve f(x) < g(x)
x – 3 < 2x + 4
-2x - 2x
-1x – 3 < 4
+3
+3
-1x < 7 (divide both sides by -1 to solve for x
I need to switch sign because of this)
Answer x = -7
Answer x > -7
e) graph f(x) and g(x) on the same graph, and label the solution to the equation f(x) = g(x)
for f(x) plot -3 on y-axis then go up 1 right 1
for g(x) plot 4 on y-axis then go up 2 right 1
the x –coordinate of the intersection point is the answer to part c x = -7
get the y-coordinate of the intersection point by plugging x = -7 into either function
f(-7) = -7 – 3 = -10
15) A jeweler’s salary was $30,000 in 2009 and $33,000 in 2012. The jeweler’s salary follows a linear
growth pattern.
a) Create a linear function to model the data
b) Use the function to estimate the jeweler’s salary in 2014.
a) First create two points I will make my points of the form (year, salary)
(2009, 30000) and (2012, 33000)
Now find the desired equation:
Find m 𝑚 =
33000−30000
2012−2009
=
3000
3
m = 1000
now find b
use y = mx + b replace m = 1000, x = 2009 y = 30000
30000 = 1000(2009) + b
30000 = 2,009,000 + b
-1,979,000 = b
desired equation y = 1000x – 1,979,000
I will give the equation a better name since the problem asked for a function
Answer s(x) = 1000x – 1,979,000
b) just plug in 2014 for x
s(2014) = 1000*2014 – 1,979,000
salary in 2014 will be $35,000
(the s stands for salary)
17) A sub shop purchases a used oven for $1,000. After 5 years the oven will need to be replaced, and
will have a value of $200. Assume the value of the oven goes down by the same amount each year.
a) Write a linear function giving the value of the equipment for the 5 years it will be in use.
Create two points my points will be in the form (year, value)
(0, 1000) (5, 200)
Now find m
𝑚=
200−1000
5−0
=
−800
5
m = -160
now find b
use y =mx + b let m = -160 x = 0 y = 1000
1000 = -160*0+b
1000 = b
Answer V(x) = -160x + 1000 ( I used a function name V for value instead of the letter y)
b) Use the function to find the value of the equipment after 3 years.
Just put in 3 for x
V(3) = -160*3 + 1000
Answer the oven will be worth $520
19) A real estate office manages an apartment complex with 100 units. When the rent is $600 per
month, all 100 units will be occupied. However, when the rent is $700 per month only 90 units will be
occupied. Assume the relationship between the monthly rent and the number of units occupied is
linear.
a) write the equation giving the demand y in terms of the price p
create two points, the hint wants you to use points in the form (price, demand)
(600, 100) (700, 90)
90−100
Find m 𝑚 = 700−600 =
−10
100
=
−1
10
Now find b: I will use m = -1/10 x = 600 y = 100 in the equation y = mx+b
100 =
−1
∗
10
600 + 𝑏 (I multiplied the fraction with my calculator)
100 = -60 + b
160 = b
−1
Answer D(p) = 10 𝑝 + 160 (I gave a fancy name, but you could use f(x), in my name the D stands for
demand and p the price)
b) Use the equation to predict the number of units occupied when the price is $675. (round to the
nearest unit if needed)
plug in 675 for p
−1
D(675) = 10 ∗ 675 + 160
= 92.5
I will round up to 93 units
c) What price would cause 80 units to be occupied?
80 =
−1
𝑝
10
+ 160
−1
-10*80 = −10 ∗ 10 𝑝 + −10 ∗ 160
-800 = p – 1600
800 = p ( a price of $800 will cause 80 units to be occupied)
25) A truck rental company rents a truck for $50 per day plus 40 cents per mile.
a) Write a linear model that relates the Cost C, in dollars of renting the truck for one day when x miles
are driven.
I can write this equation using logic and can create the equation without any algebra
C(x) = 50 + 0.40x
b) What is the cost of renting the truck for a day when 100 miles are driven?
C(100) = 50 + 0.40*100
= 50 + 40
answer Cost = $90
c) Assume the cost of renting the truck is $130. How many miles were driven?
130 = 50 + 0.40x
-50 -50
80 = 0.40x
200 = x
answer 200 miles were driven
27) A phone company offers a monthly domestic long distance package by charging $10 plus 5 cents per
minute.
a) Write a linear function that models the cost in dollars to the number of minutes used.
I can write this equation using logic and can create the equation without any algebra
C(x) = 10 + .0.05x
b) Use the function to compute the monthly cost when 500 minutes are used.
C(500) = 10 + 0.05*500
Answer cost = $35
c) How many minutes were used if the monthly cost was $40.00?
40 = 10 + 0.05x
-10 -10
30 = 0.05x
600 = x
answer 600 minutes