Math 135
Linear Functions
Solutions
Find the linear functions satisfying the given conditions:
1. f (3) = 2 and f (−3) = −4.
We have the points (3, 2) and (−3, −4). The slope of the line connecting these two
points is
a=
y2 − y1
2 − (−4)
6
=
= = 1.
x2 − x1
3 − (−3)
6
We may then input one of the points to calculate b for our linear function f (x) =
ax + b.
2 = 1(3) + b ⇒ 2 = 3 + b ⇒ b = −1,
so f (x) = x − 1, and we check our work
f (−3) = −3 − 1 = −4.
2. f (0) = 0 and f (1) =
√
2.
√
We have the points (0, 0) and (1, 2). The slope of the line connecting these two
points is:
√
y2 − y1
2−0 √
a=
= 2
=
x2 − x1
1−0
We may then input one of the points to calculate b for our linear function f (x) =
ax + b.
and so f (x) =
√
0=
√
2(0) + b ⇒ 0 = 0 + b ⇒ b = 0 ,
2x. We check our work,
√
√
f (1) = 2(1) = 2.
3. g(2) = 1 and the graph of g is parallel to the line 6x − 3y = 2.
6x − 3y = 2 ⇒ 3y = 6x − 2 ⇒ y = 2x −
a = 2.
2
3
and parallel lines have the same slope, so
We then have the point (2, 1) to input: 1 = 2(2) + b ⇒ 1 = 4 + b ⇒ b = −3 for a final
answer of f (x) = 2x − 3.
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Math 135
Linear Functions
Solutions
4. g(2) = 1 and the graph of g is perpendicular to the line 6x − 3y = 2.
Perpendicular lines have slopes whose product is −1. The slope of the line is 2 (from
the preceding problem), so we have: m1 m2 = −1 ⇒ 2a = −1 ⇒ a = − 21
We then have the point (2, 1) to input: 1 = − 12 (2) + b ⇒ 1 = −1 + b ⇒ b = 2 for a
final answer of f (x) = − 21 x + 2.
5. The x and y-intercepts of f are 5 and −1, respectively.
We have the points (5, 0) and (0, −1). The slope of the line connecting these two
points is:
a=
0 − (−1)
1
y2 − y1
=
=
x 2 − x1
5−0
5
We may then input one of the points to calculate b for our linear function f (x) =
ax + b, to obtain
1
0 = (5) + b ⇒ 0 = 1 + b ⇒ b = −1,
5
1
so f (x) = 5 x − 1, and we check our work
1
f (0) = (0) − 1 = −1 .
5
6. During the 1990s the percentage of T V households viewing cable and satellite TV
programs increased while the percentage viewing network affiliate shows generally
decreased. The primetime ratings for the network affiliates in 1993 was 40.9. The
ratings fell in 1995 to 37.3.
(a) Find the equation for the linear function (x =year, y = f (x) =rating) whose
graph passes through the two points given.
a=
37.3 − 40.9
−3.6
9
y2 − y1
=
=
= −1.8 = −
x 2 − x1
1995 − 1993
2
5
We may then input one of the points to calculate b for our linear function f (x) =
ax + b.
9
9
37.3 = − (1995) + b ⇒ 37.3 = −3591 + b ⇒ b = 3628.3 so f (x) = − x + 3628.3
5
5
and doublecheck
9
f (1993) = − (1993) + 3628.3 = −3587.4 + 3628.3 = 40.9 .
5
(b) What should the ratings be in 2000?
9
f (2000) = − (2000) + 3628.3 = −3600 + 3628.3 = 28.3 .
5
University of Hawai‘i at Mānoa
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R Spring - 2014
Math 135
Linear Functions
Solutions
(c) What year would we expect the ratings to reach 30?
This amounts to solving the equation
9
9
30 = − x + 3628.3 ⇒ x = 3598.3 ⇒ x = 1999(approximately).
5
5
7. The population of Florida was 11,350,000 in 1985 and reached 13,000,000 in 1990.
(a) Find the equation for the linear function (x =year, y = f (x) =population)
whose graph passes through the two points given.
a=
13, 000, 000 − 11, 350, 000
1, 650, 000
y2 − y1
=
=
= 330, 000
x2 − x1
1990 − 1985
5
We may then input one of the points to calculate b for our linear function f (x) =
ax + b.
13, 000, 000 = 330, 000(1990) + b ⇒ 13, 000, 000 = 656, 700, 000 + b
⇒ b = −643, 700, 000 so f (x) = 330, 000x − 643, 700, 000
and doublecheck f (1985) = 330, 000(1985) − 643, 700, 000 = 655, 050, 000 −
643, 700, 000 = 11, 350, 000X
(b) What should the population be in 2000?
f (2000) = 330, 000(2000)−643, 700, 000 = 660, 000, 000−643, 700, 000 = 16, 300, 000
(c) What year would we expect the population to reach 25,000,000?
This amounts to solving the equation: 25, 000, 000 = 330, 000x − 643, 700, 000
⇒ 330, 000x = 668, 700, 000 ⇒ x = 2026 (approximately).
University of Hawai‘i at Mānoa
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R Spring - 2014