The exponential decay graph shows the expected depreciation for a new boat, selling for $3500, over 10 years.
A coordinate graph is shown. The horizontal axis extends from 0 to 12 years. The vertical axis extends from 0 to 9500 with an axis label of 'Value' with the dollars sign in parenthesis ($). A curve is graphed which begins at the given numbers of (0, 3500), then decreases passing through approximately (12, 100), although, this number is not officially given.
a. Write an exponential function for the graph.
b. Use the function to find the value of the boat after 9.5 years.
I would appreciate some help to solve this problem. I am about 70% sure that I know the formula that I have to use, but I came up with two different ones, and it doesn't really look right to me.
The formula that I have been using is 3500=a*e^k*0 and/or y=3500*e^k*9.5
To write an exponential function for the given graph, we can use the general form of an exponential decay function:
y = a * e^(-kx)
Where:
- y is the value of the boat at a given time (in dollars)
- a is the initial value of the boat (in dollars)
- e is the base of the natural logarithm (approximately 2.71828)
- k is the decay constant (a positive value)
In this case, we know that at time 0, the boat has an initial value of $3500, so a = 3500. We also know that the curve passes through approximately (12, 100). It means that after 12 years, the boat is expected to have a value of $100.
Plugging these values into the equation, we have:
100 = 3500 * e^(-12k)
Now, to find the value of the boat after 9.5 years, we substitute x = 9.5 into the equation:
y = 3500 * e^(-9.5k)
To solve these equations, we need to find the value of k. Unfortunately, without the specific value for (12, 100), it is challenging to determine an exact value of k or solve for the function algebraically.
However, we can use numerical methods or trial-and-error to estimate the value of k and then use that estimate to find the value of the boat after 9.5 years.
You have the function y = 3500 e^-(kx)
and it appears that
3500 e^(-12k) = 100
so k = -ln35/12
so now find y(9.5)
When you have k = -ln35/12, does the / mean divided by or is that a fraction?
ln(1/35)/12 = -ln35 / 12
which you could have determined by solving the equation ...