Question

Gerardo purchased a new boat in 2011 for $35,000. In 2013, his boat was worth $27,300.Let the linear function, f, determine the value (in dollars) of Gerardo's boat in terms of the number of years, t, since 2011. Find the function f.f(t)=Next, let the exponential function, g, determine the value (in dollars) of Gerardo'ss boat in terms of the number of years, t, since 2011. Find the function g.g(t)=

Answer 1

Given:

The cost of boat in 2011 is $35,000.

The cost of boat in 2012 is $27,300.

The objective is to find,

a) The linear function f(t).

a) The exponential function g(t).

Step-by-step explanation:

Consider the given points as,

`\begin{gathered} (x_1,y_1)=(2011,35000) \\ (x_2,y_2)=(2012,27300) \end{gathered}`

a)

The linear equation using two points can be calculated as,

`y-y_1=((y_2-y_1))/((x_2-x_1))(x-x_1)\text{ . . . . . .(1)}`

On plugging the coordinates in equation (1),

`\begin{gathered} y-35000=((27300-35000))/((2013-2011))(x-2011) \\ y-35000=-3850(x-2011) \\ y-3500=-3850x+7742350 \\ y=-3850x+7742350+3500 \\ y=-3850x+7777350 \end{gathered}`

Hence, the linear function is f(t) = -3850x+7777350.

b)

The exponential function can be calculated as,

`y=ab^x\text{ . . . . .(1)}`

First substitute (x1,y1) in equation (1).

`35000=ab^(2011)\text{ . . . . . . (2)}`

Now, substitute (x2,y2) in equation (2),

`27300=ab^(2013)\text{ . . . . . . .(3)}`

To find b:

Divide the equations (2) and (3).

`\begin{gathered} (35000)/(27300)=(ab^(2011))/(ab^(2013)) \\ 1.282=b^(2011-2013) \\ 1.282=b^(-2) \\ b=1.282^{-(1)/(2)} \\ b=0.883 \end{gathered}`

To find a:

Now, substitute the value of b in equation (2).

`\begin{gathered} 35000=a(0.883)^(2011) \\ a=(35000)/(0.883^(2011)) \end{gathered}`