Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Understanding Relations and Functions
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Exercise 11 Page 104

Remember that a function is a type of relation for which there is exactly one output for each input.

Table:

Number of hours Fee
1 175
2 175
3 225
4 275

Domain: {1,2,3,4}
Range: {175,225,275}
Interpretation of the Context: See solution.
Relation: The relation is a function.

Practice makes perfect
We have been told that an electrician charges a base fee of $75 plus $50 for each hour of work. Let x be the number of hours and y be the total cost. We can write the relation between x and y as an equation. y= 50x + 75 Let's make table of values for the situation.

Table

Remember that the minimum the electrician charges is $175.

x Relation: y=50x+75 y
1 y=50( 1)+75 125→ 175
2 y=50( 2)+75 175
3 y=50( 3)+75 225
4 y=50( 4)+75 275

Domain and Range

Let's determine the domain and the range of the relation. First, let's recall what the domain and range of a relation are.

  • The domain is the possible inputs of a relation.
  • The range is the possible outputs of a relation.

In this case, the number of hours will be the domain and the total cost will be the range. Domain:& {1,2,3,4} Range:& {175,225,275}

Interpretation of the Context

Let's interpret the context of these results.

  • For an input of 1 hour there is one output of $175.
  • For an input of 2 hours there is one output of $175.
  • For an input of 3 hours there is one output of $225.
  • For an input of 4 hours there is one outputs of $275.

Is the relation a function?

A function is a type of relation in which there is exactly one output for each input. In the opposite direction, it does not matter how many inputs give the same output. Therefore, our relation is a function.