Question

Depreciation is the decrease or loss in value of an item due to age, wear, or market conditions. We usually consider depreciation on expensive items like cars. Businesses use depreciation as a loss when calculating their income and taxes.

One company buys a new bulldozer for $93350. The company depreciates the bulldozer linearly over its useful life of 15 years. Its salvage value at the end of 15 years is $16100.

A) Express the value of the bulldozer, V, as a function of how many years old it is, t. Make sure to use function notation.


B) The value of the bulldozer after 12 years is $

Answer 1

we know the depreciation is linear, so what we have is a linear equation for it, we also know that the bulldozer when new is $93350, that is, at year "0" is that much, and at year 15 is $16100, so expressing its value V as a function of "x", so we can say that the value is the "dependent" variable and thus "x" the "independent" one, so we can express V over the y-axis whilst "x" well, that's obvious :).

to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below.

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{93350})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{16100}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{16100}-\stackrel{y1}{93350}}}{\underset{\textit{\large run}} {\underset{x_2}{15}-\underset{x_1}{0}}} \implies \cfrac{ -77250 }{ 15 } \implies - 5150`

`\begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{93350}=\stackrel{m}{- 5150}(x-\stackrel{x_1}{0}) \\\\\\ y=-5150x+93350\implies \boxed{V(x)=-5150x+93350} \\\\\\ \stackrel{\textit{after 12 years}}{V(12)=-5150(12)+93350}\implies \boxed{V(12)=31550}`