Does each function describe exponential growth or decay?

Drag and drop the equations into the boxes to correctly complete the table.


Growth Decay
y=100(1−12)^t


y=0.1(1.25)^t

y=((1−0.03)12)^2t

y=426(0.98)^t

y=2050(12)^t

A function [tex]f[/tex] from a set [tex]A[/tex] to a set [tex]B[/tex] is a relation that assigns to each element [tex]x[/tex] in the set [tex]A[/tex] exactly one element [tex]y[/tex] in the set [tex]B[/tex]. The set [tex]A[/tex] is the domain (also called the set of inputs) of the function and the set [tex]B[/tex] contains the range (also called the set of outputs).

[tex]We \ denote \ the \ \mathbf{exponential \ function} \ f \ with \ base \ a \ as: \\ \\ f(x)=a^x \\ \\ where \ a>0, \ a\neq 1, \ and \ x \ is \ any \ real \ number[/tex].

We have the following equation:

[tex]y=100(1-12)^t[/tex]

That can be written as:

[tex]y=100(-11)^t[/tex]

Recall that the definition of exponential functions establishes that:

[tex]We \ denote \ the \ \mathbf{exponential \ function} \ f \ with \ base \ a \ as: \\ \\ f(x)=a^x \\ \\ where \ a>0, \ a\neq 1, \ and \ x \ is \ any \ real \ number[/tex].

That is:

[tex]a \ \mathbf{must} \ be \ greater \ than \ 1[/tex]

In this problem, [tex]a=-11[/tex], therefore this is not an exponential function.

The function:

[tex]y=0.1(1.25)^t[/tex]

is an exponential function because is a function of the form [tex]f(t)=ka^t \\ \\ where \ a>0 \ and \ k \ constant[/tex]

So [tex]k=0.1 \ and \ a=1.25[/tex]. Since [tex]a \ is \ greater \ than \ 1[/tex] and being raised to the power of [tex]t[/tex], the function increases. This means that [tex]y[/tex] increases as [tex]t[/tex] increases as illustrated in Figure 1. This represents a growth.

The function:

[tex]y=((1-0.03)12)^{2t}[/tex] can be written as:

[tex]y=11.64^{2t}[/tex]

and is an exponential function because is a function of the form [tex]f(t)=a^{bt} \\ \\ where \ a>0 \ and \ b \ constant[/tex]

So [tex]a=11.64 \ and \ b=2[/tex]. Since [tex]a \ is \ greater \ than \ 1[/tex] and being raised to the power of [tex]2t[/tex], the function increases. As in the previous exercise, this means that [tex]y[/tex] increases as [tex]t[/tex] increases as illustrated in Figure 2. This represents a growth.

The function:

[tex]y=426(0.98)^t[/tex]

is an exponential function because is a function of the form [tex]f(t)=ka^t \\ \\ where \ a>0 \ and \ k \ constant[/tex]

So [tex]k=426 \ and \ a=0.98[/tex]. Since [tex]a \ is \ less \ than \ 1[/tex] and being raised to the power of [tex]t[/tex], the function decreases. Here this means that [tex]y[/tex] decreases as [tex]t[/tex] increases as illustrated in Figure 3. This represents a decay.

The function:

[tex]y=2050(12)^t[/tex]

is an exponential function because is a function of the form [tex]f(t)=ka^t \\ \\ where \ a>0 \ and \ k \ constant[/tex]

So [tex]k=2050 \ and \ a=12[/tex]. Since [tex]a \ is \ greater \ than \ 1[/tex] and being raised to the power of [tex]t[/tex], the function increases. So in this function [tex]y[/tex] also increases as [tex]t[/tex] increases as illustrated in Figure 4. This represents a growth.

Answer:

1. isn't an exponential function

2. growths

3. growths

4. decay

5. growths

Step-by-step explanation:

The exponential functions has the form: y = k1*k2^(k3*t), where k1, k2 and k3 are a constants, and k2>0. Graph of exponential functions always decrease or always increase. To know if a function growths or decay just evaluate the function in 2 points, for example t = 0 and t = 1, and compare their results.

1.

y = 100*(1−12)^x = 100*(−11)^x

Since (-11) is negative, then y is not an exponential function

2.

y(0) = 0.1*(1.25)^(0) = 0.1

y(1) = 0.1*(1.25)^(1) = 0.125

y(1) > y(0) -> growths

3.

y(0) =((1−0.03)12)^2(0) = 1

y(0) =((1−0.03)12)^2(1) = 135.4896

y(1) > y(0) -> growths

4.

y(0) = 426(0.98)^(0) = 426

y(1) = 426(0.98)^(1) = 417.48

y(1) < y(0) -> decay

5.

y(0) = 2050(12)^(0) = 2050

y(1) = 2050(12)^(1) = 24600

y(1) > y(0) -> growths

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