Big Ideas Math Algebra 1, 2015
BI
Big Ideas Math Algebra 1, 2015 View details
4. Exponential Growth and Decay
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Exercise 72 Page 322

Practice makes perfect
Recall that an exponential function f(t)=ab^t, with a>0, exhibits exponential growth if b>1. Therefore, we can compare our function to this form and verify that this requirement is satisfied.
a 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

As we can see, r= 1. This means that our function increases by 100 %, or doubles, each time t increases by 1.
b 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.
g(t) = kf(t)
g(t) = k 2^t
g( 0) = k2^0
4 = k2^0
â–Ľ
Solve for k
4 = k* 1
4 = k
k = 4
Therefore, g(t)=4(2)^t.
c 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).Then, we need to write g(t)=4(2)^t in the form 2^(t+r), so this way we can compare them and find the value of r. We can start to rewrite g(r) by writing 4 as 2^2. Then we can simplify it by using the Product of a Power Property.
g(t)=4(2)^t
g(t)=2^2 * 2^t
g(t)=2^(2+t)
g(t)=2^(t+2)
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.