Algebra II - Graphing Exponential Functions

Determine whether each function represents exponential decay or growth.

$ewline a) ewline y=(\frac{1}{2})^{x}$

a) decay

b) growth

a) growth

b) growth

a) decay

b) decay

a) growth

b) decay

Explanation

This is exponential decay since the base, , is between and .

This is exponential growth since the base, , is greater than .

Match each function with its graph.

a. 3time2tothex

b. 1over2tothex

c. 2_tothe_x

Explanation

For , our base is greater than so we have exponential growth, meaning the function is increasing. Also, when , we know that since . The only graph that fits these conditions is .

For , we have exponential growth again but when , . This is shown on graph .

For , we have exponential decay so the graph must be decreasing. Also, when , . This is shown on graph .

What is the -intercept of the graph $y = 3^{x}$ ?

$\frac{1}{3}$

Explanation

The -intercept of any graph describes the -value of the point on the graph with a -value of .

Thus, to find the -intercept substitute .

In this case, you will get,

$y = 3^{0} = 1$ .

In 2010, the population of trout in a lake was 416. It has increased to 521 in 2015.

Write an exponential function of the form that could be used to model the fish population of the lake. Write the function in terms of , the number of years since 2010.

Explanation

We need to determine the constants and . Since in 2010 (when ), then and

To get , we find that when , . Then $b^5=\frac{521}{416}$ .

Using a calculator, ${\left(\frac{521}{416}\right)}^{\frac{1}{5}}=1.046$ , so $b\approx 1.046$ .

Then our model equation for the fish population is

What is the -intercept of $y = 5^{x+2}$ ?

$\frac{1}{25}$

Explanation

The -intercept of any function describes the point where .

Substituting this in to our funciton, we get:

$y = 5^{x+2} = 5^{0+2} = 5^{2} = 25$

What is the -intercept of $y = 4(2^{x})$ ?

There is no -intercept.

Explanation

The -intercept of a graph is the point on the graph where the -value is .

Thus, to find the -intercept, substitute and solve for .

Thus, we get:

$y = 4(2^{0}) = 4(1)=4$

Which of the following correctly describes the graph of an exponential function with a base of three?

It starts out by gradually increasing and then increases faster and faster.

It starts by increasing quickly and then levels out.

It begins by decreasing quickly and then levels out.

It begins by decreasing gradually and then decreases more quickly.

It stays constant.

Explanation

Exponential functions with a base greater than one are models of exponential growth. Thus, we know that our function will increase and not decrease. Remembering the graph of an exponential function, we can determine that the graph will begin gradually, almost like a flat line. Then, as increases, begins to increase very quickly.

An exponential funtion is graphed on the figure below to model some data that shows exponential decay. At , is at half of its initial value (value when ). Find the exponential equation of the form $y=Ae^{kt}$ that fits the data in the graph, i.e. find the constants and .

Expdecay

$y=1.20e^{-1.9t}$

$y=0.3648e^{-0.6t}$

$y=0.3648e^{0.6t}$

$y=2.4e^{-0.6t}$

Explanation

To determine the constant , we look at the graph to find the initial value of , (when ) and find it to be . We can then plug this into our equation $y=Ae^{kt}$ and we get $1.2=Ae^{k0}$ . Since $e^{k0}=e^0=1$ , we find that .

To find , we use the fact that when , is one half of the initial value . Plugging this into our equation with now known gives us $\frac{1.2}{2}=0.6=1.2e^{k(0.3648)}$ . To solve for , we make use the fact that the natural log is the inverse function of , so that

$\ln(e^{k(0.3648)})=k(0.3648)$ .

We can write our equation as $\frac{0.6}{1.2}=e^{k(0.3648)}$ and take the natural log of both sides to get:

$\ln(\frac{0.6}{1.2})=\ln(e^{k(0.3648)})$ or .

Then $k\approx -1.9$ .

Our model equation is $y=1.20e^{-1.9t}$ .

Give the -intercept of the graph of the equation $f(x) = 2 ^{x- 7} + 4$ .

The graph has no -intercept.

Explanation

Set and solve for

$f(x) = 2 ^{x- 7} + 4$

$2 ^{x- 7} + 4 = 0$

$2 ^{x- 7} = -4$

We need not work further. It is impossible to raise a positive number 2 to any real power to obtain a negative number. Therefore, the equation has no solution, and the graph of has no -intercept.