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).

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,

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

On plugging the coordinates in equation (1),

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

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).

$\begin{gathered} \frac{35000}{27300}=\frac{ab^{2011}}{ab^{2013}} \\ 1.282=b^{2011-2013} \\ 1.282=b^{-2} \\ b=1.282^{-\frac{1}{2}} \\ b=0.883 \end{gathered}$

To find a:

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

$\begin{gathered} 35000=a(0.883)^{2011} \\ a=\frac{35000}{0.883^{2011}} \end{gathered}$