Question 628699
First, this problem has nothing to do with Trigonometry. It belongs in one of the categories related to exponents. Please post using a possibly relevant category.<br>
Second, without the graph you were given it will be impossible for a tutor to solve this problem. I will, nevertheless, try to help as much as I can.<br>
The general equation for exponential growth and decay is:
{{{A = A[0]*(1+r)^t}}}
where
t is a number of units of time. (Often the units are years so t would be a number of years.)
A = the amount after "t" units of time.
{{{A[0]}}} is the starting amount (IOW the amount at t = 0. This is why the A has a subscript of zero.)
r is a decimal (or fraction) between 0 and 1 which represents the rate of change. A positive r means growth. A negative r means decay. (Note: Some books use {{{A = A[0]*r^t}}}, merging the 1 and r. In this case if r > 1 then the equation is for growth and if r is between 0 and 1 then the equation is for decay. I prefer the first form I gave you because that percent growths or decays are easier to use with that form.)<br>
You have been given the original amount, {{{A[0]}}}. So your equation so far is:
{{{A = 3500*(1+r)^t}}}<br>
From this point on I can only tell you what you should do to finish:<ol><li>Find the value for r:<ol><li>Look at the graph and find the coordinates of any point (other than where t = 0) on the graph. (If the axes are not labeled t and A, then the "t" in the equation corresponds to "x" and the "A" corresponds to the "y".)</li><li>Replace the t and A in your equation with the coordinates you found in step 1.</li><li>Solve this equation for r.</li></ol></li><li>Rewrite your working equation:
{{{A = 3500*(1+r)^t}}}
substituting in the value for r you just found. This (or the equation you get if you add the 1 to r) is the answer to part a.</li><li>Using the equation from part a, replace the t with 9.5 and solve for A. This will be the answer for part b.</li></ol>
Here's an example:
1. Solve for r
1.1 Find a point. Suppose that your graph shows that at t (or x) = 10, the A (or y) is 700.
1.2 Insert the coordinates into your working equation:
{{{700 = 3500*(1+r)^10}}} 
1.3. Solve for r.
Dividing both sides by 3500:
{{{0.2 = (1+r)^10}}} 
Find the 10th root of each side. (On your calculator use 0.2^(1/10) or 0.2^0.1)
{{{0.85133992 = 1 + r}}}
Subtract 1 from each side:
{{{-0.14866008 = r}}}
(BTW: As a percent, r is an approximately 15% rate of decay.)<br>
2. Replace the r in your working equation:
{{{A = 3500*(1+-0.14866008)^t}}}
This, or {{{A = 3500*(0.85133992)^t}}}, would be the answer to part a.<br>
3. To find the value of the car after 9.5, replace the t in you equation with 9.5 and solve for A:
{{{A = 3500*(1+-0.14866008)^(9.5)}}}
Simplifying...
{{{A = 3500*(0.85133992)^(9.5)}}}
Using 0.85133992^(9.5) on our calculators:
{{{A = 3500*0.21675967}}}
{{{A = 758.65884937}}}
So after 9.5 years the car would be worth approximately $758.69.