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An exponential growth function can be written in the form f(x) = a(1+r)^x.
Look at the graph carefully. How do the functions' values compare? Is k>1 or 0
Find the explicit form of f(t+r). Then, try to write g(t) in this form to identify r.
Is it an exponential growth function? Yes.
Explanation: See solution.
Rate of growth: 1, or 100 %
What kind of transformation? It is a vertical stretch.
Explanation: See solution.
See solution.
f(t)= a b^t
f(t)= 1( 2)^t
As we can see, a= 1 and b= 2. Therefore, since a>0 and b>1, this is an exponential growth function. Now that we know this is the case, we can write it in the form f(t) = a(1+r)^t.
f(t) = 2^t ⇔ f(t) = 1(1+1)^t
Finally, we can identify the rate of growth r by direct comparison.
f(t) = a(1+ r)^t
f(t) = 1(1+ 1)^t
Notice that g(t) equals f(t) times a constant value k. Let's take look at the graph.
As we can see, the values of g(t) are greater than those of f(t) for all x-values. Therefore k must be greater than 1, and consequently, g(t) is a vertical stretch of f(t). To find the value of k, we can substitute a point and the explicit form of f(t). For example, notice that g(0) = 4. Let's substitute this information.
f(t)= 2^t
t= 0
g(0)= 4
Therefore, g(t)=4(2)^t.
From Part B we know that g(t)=4(2)t. In order to compare it with h(t), we need to write it in the form f(t+r).
f(t) = 2^t ⇒ f(t+r) = 2^(t+r).
Rewrite 4 as 2^2
a^m*a^n=a^(m+n)
Commutative Property of Addition
Now that g(t) is written in this form, we can compare it directly with h(t) = 2^(t+r) and identify r. h(t) = 2^(t+ r) g(t)=2^(t+ 2) As we can see, r= 2.