Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Using Intercepts
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Exercise 23 Page 178

Practice makes perfect
a We are told that Kathryn needs 30 minutes to complete 2-mile walk. With this information we can calculate her pace.

distance/time=2/30=1/15

Let's write a function f(x) where x represents the number of minutes that have passed and f(x) is the remaining distance. Therefore, in the beginning the distance is 2 miles and decreases by the pace multiplied by the number of minutes that have passed.
f(x)= 2-1/15x

Next, we will graph the function by creating a table of values.

x 2-1/15x f(x)
6 2-1/15( 6) 1.6
12 2-1/15( 12) 1.2
18 2-1/15( 18) 0.8
24 2-1/15( 24) 0.4

Now, let's plot the points and draw the function.

b In this part, we will find and interpret the interception points. Let's find the x-intercept first. To do so, we will substitute 0 for f(x).
f(x)=2-1/15x
0=2-1/15x
1/15x=2
x=30
The x-intercept is 30. This means that, after 30 minutes, the distance remaining will be 0. Now, we will find the y-intercept by substituting 0 for x.
f(x)=2-1/15x
f( 0)=2-1/15( 0)
f(0)=2
The y-intercept is 2. This means that the distance at the beginning is 2 miles.