McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
6. Analyzing Functions with Successive Differences
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Exercise 28 Page 594

Practice makes perfect
a The table below shows the relationship between the cost and the length of the call.

Year 0 1 2 3 4
Value ($) 18 500 15 910 13 682.60 11 767.04 10 119.65

Let's plot the ordered pairs ( 0, 18 500), ( 1, 15 910), ( 2, 13 682.60), ( 3, 11 767.04), and ( 4, 10 119.65).

The ordered pairs appear to represent an exponential function.

Extra

Linear and Nonlinear Functions
The general forms of the equations and a graph each function type are listed below.

b

We can model the value y of the car after x years by using an exponential function. y=ab^xHere, a is the initial value and b is the constant multiplier. From the table, we see that the initial value is 18 500. Let's place it into the function. y= ab^x ⇔ y= 18 500(b)^x We can find b by substituting an ordered pair from the given table. Let's substitute (1,15 910).

y=18 500(b)^x
15 910=18 500(b)^1
â–¼
Solve for b
15 910=18 500b
15 910/18 500=b
0.86=b
b=0.86

The constant multiplier is 0.86. Then, the function below models the data. y=18 500( b)^x ⇓ y=18 500( 0.86)^x

c Using the model we found in Part B, we can find the value of the car after 7 years. To do it, we need to substitute 7 for x.

y=18 500(0.86)^x
y=18 500(0.86)^7
â–¼
Simplify right-hand side
y=18 500(0.34792 ...)
y=6436.66471 ...
y=6436.66

Therefore, the car is worth about $6436.66 after 7 years.