The exponential decay chart shows the expected depreciation for a new boat that sells for $3,500 over 10 years a. write an exponential function for the graph. B use the function from part a to find the value of the boat after 9.5 years.
A = Ao x (1+r)^t wheret = number of time units. A = the amount after “t” time units. Ao = The initial amount Hence the final equation after substitution: A = 3500 x(1+r)^10
(a) (b) $245.27 Explanation: (a) From the graph below, it is clear that the graph passes through the points (0.3500) and (2.2000). The general form of an exponential function is where, a is the initial value and b is the growth or decay factor. The initial value is 3500, it means a=3500. f(x)=2000 to x=2. The exponential function of the graph is (b) We need to find the value of the boat after 9.5 years. Substitute x=9.5 in the above function. Therefore, the value of the boat after 9.5 years is $245.27.
and
=
3500
e^(-k⋅
9.5)
Walkthrough: The exponential decay chart shows the expected depreciation for a new boat, sold for $3,500, over 10 years a. Write an exponential function for the graph.
B Use the function in part a to find the value of the boat after 9.5 years. Explanation:
The exponential equation is given by
and
=
3500
⋅
e ^ (
k
9.5)
by which:
and
: assess
A
: constant;
k
: rate of change
you
: time value In this when t =0 3500=
A
⋅
e ^
k
0
3500
=
A
after 10 years we have, after 9.5 years the value of the boat is:
and
=
3500
e^(-k⋅
9.5)
k is the rate of change and shows that it is negative because there is depreciation in value. Note that the rate of change is not given in this case.
n = 0 V(0) = a * b^0 = 3500 a = 3500 V(2) = a * b^2 2000 = 3500 * b^2 b = sqrt (2000/3500) b ≈ 0.76 V( n) = 3500 * 0.76^n We can check for n = 1 that it is close to 2500 on the graph: V(1) = 3500 * (0.76)^1 V(1) = 2660 And on the graph we have V( 3 ) ≈ 1500, V(n) = 3500 * (0.76)^3 ≈ 1536 Now n = 9.5 V(9.5) = 3500 * (0.76)^(9.5) V (9,5) ≈ 258