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SAT II Math I : Solving Exponential Functions

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #21 : Solving Exponential Functions

What is/are the asymptote(s) of the graph of the function $g(x) = 0.4 ^{x - 2} + 0.7$ ?

Possible Answers:

y = 0.4

y = 1.1

y = 0.7

x = 2

x = 2 and y = 0.7

Correct answer:

y = 0.7

Explanation:

An exponential function of the form

$g(x) = a^{x- h}+k$

has as its one and only asymptote the horizontal line y = k .

Since we define as

$g(x) = 0.4 ^{x - 2} + 0.7$ ,

then k= 0.7 ,

and the only asymptote is the line of the equation y = 0.7 .

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Example Question #21 : Solving Exponential Functions

Determine whether each function represents exponential decay or growth.

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

$\newline b) \newline y=0.4(1.1)^x$

Possible Answers:

a) growth

b) decay

a) decay

b) growth

a) growth

b) growth

a) decay

b) decay

Correct answer:

a) decay

b) growth

Explanation:

This is exponential decay since the base, (1/2) , is between and .

This is exponential growth since the base, 1.1 , is greater than .

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Example Question #4 : Graphing Exponential Functions

Match each function with its graph.

1. y=2^x

2. y=3(2^x)

3. y=(1/2)^x

a. 3time2tothex

b. 1over2tothex

c. 2_tothe_x

Possible Answers:

Correct answer:

Explanation:

For y=2^x , our base is greater than so we have exponential growth, meaning the function is increasing. Also, when x=0 , we know that y=1 since 2^0=1 . The only graph that fits these conditions is .

For y=3(2^x) , we have exponential growth again but when x=0 , y=3 . This is shown on graph .

For y=(1/2)^x , we have exponential decay so the graph must be decreasing. Also, when x=0 , y=1 . This is shown on graph .

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Example Question #71 : Functions And Graphs

An exponential funtion y(t) is graphed on the figure below to model some data that shows exponential decay. At t=0.3648 , is at half of its initial value (value when t=0 ). Find the exponential equation of the form $y=Ae^{kt}$ that fits the data in the graph, i.e. find the constants and .

Expdecay

Possible Answers:

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

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

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

$y=0.3648e^{0.6t}$

Correct answer:

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

Explanation:

To determine the constant , we look at the graph to find the initial value of , (when t=0 ) and find it to be y(0)= 1.2 . 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 A=1.2 .

To find , we use the fact that when t=0.3648 , is one half of the initial value y(0)=1.2 . 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 -0.6931=k(0.3648) .

Then $k\approx -1.9$ .

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

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Example Question #6 : Graphing Exponential Functions

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 y=ab^x that could be used to model the fish population of the lake. Write the function in terms of , the number of years since 2010.

Possible Answers:

y=1.2(416)^x

y=416(521)^x

y=416(1.046)^x

y=521(1.005)^x

y=1.005(521)^x

Correct answer:

y=416(1.046)^x

Explanation:

We need to determine the constants and . Since y=416 in 2010 (when x=0 ), then y(0)=ab^0=a and a=416

To get , we find that when x=5 , y(5)=416b^5=521 . 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 y=416(1.046)^x

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Example Question #21 : Solving Exponential Functions

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

Possible Answers:

$\frac{1}{3}$

Correct answer:

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 x = 0 .

In this case, you will get,

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

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Example Question #2 : Graph Exponential Functions

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

Possible Answers:

There is no -intercept.

Correct answer:

Explanation:

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

Thus, to find the -intercept, substitute x = 0 and solve for .

Thus, we get:

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

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Example Question #9 : Graphing Exponential Functions

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

Possible Answers:

$\frac{1}{25}$

Correct answer:

Explanation:

The -intercept of any function describes the point where x = 0 .

Substituting this in to our funciton, we get:

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

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Example Question #21 : Solving Exponential Functions

Which of the following functions represents exponential decay?

Possible Answers:

$y = 5^{x}$

$y = 3(2^{x})$

$y = \left(\frac{1}{3}\right)^{x}$

$y = 6^{x + 3}$

$y = (-2)^{x}$

Correct answer:

$y = \left(\frac{1}{3}\right)^{x}$

Explanation:

Exponential decay describes a function that decreases by a factor every time increases by .

These can be recognizable by those functions with a base which is between and .

The general equation for exponential decay is,

A=A_0b^t where the base is represented by and 0<b<1 .

Thus, we are looking for a fractional base.

The only function that has a fractional base is,

$y = \left(\frac{1}{3}\right)^{x}$ .

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Example Question #122 : Solving And Graphing Exponential Equations

Does the function $y = 4^{x}$ have any -intercepts?

Possible Answers:

Yes, x = 1

Yes, x = 4 and x = -4

Yes, x = 0

That cannot be determined from the information given.

Correct answer:

Explanation:

The -intercept of a function is where y = 0 . Thus, we are looking for the -value which makes $0 = 4^{x}$ .

If we try to solve this equation for we get an error.

To bring the exponent down we will need to take the natural log of both sides.

ln(0)=xln(4)

Since the natural log of zero does not exist, there is no exponent which makes this equation true.

Thus, there is no -intercept for this function.

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All SAT II Math I Resources

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SAT II Math I : Solving Exponential Functions

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #21 : Solving Exponential Functions

Example Question #21 : Solving Exponential Functions

Example Question #4 : Graphing Exponential Functions

Example Question #71 : Functions And Graphs

Example Question #6 : Graphing Exponential Functions

Example Question #21 : Solving Exponential Functions

Example Question #2 : Graph Exponential Functions

Example Question #9 : Graphing Exponential Functions

Example Question #21 : Solving Exponential Functions

Example Question #122 : Solving And Graphing Exponential Equations

All SAT II Math I Resources

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