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Plot the ordered pairs. Then, determine its type.
Use the general form of the function to write an equation.
Substitute the given value into the equation you found in Part B.
Graph:
Type of Function: Exponential
Equation: y=18 500(0.86)^x
About $6436.66
The table below shows the relationship between the cost and the length of the call.
| Year | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Value ($) | 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.
We can model the value y of the car after x years by using an exponential function.
y=ab^x
x= 1, y= 15 910
a^1=a
.LHS /18 500.=.RHS /18 500.
Calculate quotient
Rearrange equation
The constant multiplier is 0.86. Then, the function below models the data. y=18 500( b)^x ⇓ y=18 500( 0.86)^x
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.
Therefore, the car is worth about $6436.66 after 7 years.