Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Formalizing Relations and Functions
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Exercise 29 Page 272

Practice makes perfect
a A theater group is having a car-wash fundraiser. We are given the following information.
  • Each car is charged $5.
  • The soap costs $34.
  • They have enough soap to wash 40 cars.
The number of cars washed c is the independent variable. It affects the profit p, which is the dependent variable. We can say that p depends on c.

b The relationship between p and c is a function because there is a unique amount of earned profit p for each number of cars c that are washed. It is not possible to make two different profits out of the same number of washed cars.

c To write an equation for this relationship, we will translate the verbal expressions into algebraic expressions. Let's make a table for this purpose.
Verbal Expression Algebraic Expression
Amount earned for one car $5
Number of cars washed c
Amount earned for c numbers of cars being washed 5c
Cost of soap $34
Amount earned from washing cars minus the initial cost of the soap 5c- 34

Therefore, the equation for the profit p can be written as p= 5c- 34.

d Since the group can wash at least 0 cars and at most 40 cars using their purchased soap, the domain consists of all integers between 0 and 40 inclusive.

0 ≤ c ≤ 40 The range can be found by substituting the limits of the domain into the function. Let's first find the lower limit of the range.

p=5c-34
p=5* 0-34
â–¼
Evaluate right-hand side
p=0-34
p=-34

If the group washes no cars, the loss will be $ 34. This means there is $0 in profit, so 0 will be our lower limit. Let's now find the upper limit for the range.

p=5c-34
p=5* 40-34
â–¼
Evaluate right-hand side
p=200-34
p=166

The upper limit 166. The theater group will make a $166 profit if they wash 40 cars. Therefore, we can write the range as follows. 0 ≤ p ≤ 166