<u>Let's first consider:</u><em> the fact that a new computer loses 1/2 of its value</em>
<em> </em> ⇒ <u>third choice</u> and <u>fourth choice</u> are <u>eliminated</u>
⇒ because of the graph project the computer's value as growing over
years
<u>Now let's try to set up an equation</u>:
After 1st year ⇒ Computer's value = 
After 2nd year ⇒ Computer's value = 
After 3rd year ⇒ Computer's value =
..... (<em>Hopefully you see a pattern)</em>
<em />
After x year ⇒ Computer's value = 
<u>Therefore, the equation is</u>: 
⇒that means the graph has to <u>exponentially decreasing</u> as it has
an exponent (<em>meaning the function is exponential</em>) and a base less
than 1 (<em>meaning that the function is decreasing)</em>
<u>Answer: Choice (1)</u>
<u></u>
Hope that helps!