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SAT II Math I : SAT Subject Test in Math I

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

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6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #361 : Sat Subject Test In Math I

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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Example Question #362 : Sat Subject Test In Math I

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

Possible Answers:

It begins by decreasing gradually and then decreases more quickly.

It begins by decreasing quickly and then levels out.

It stays constant.

It starts by increasing quickly and then levels out.

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

Correct answer:

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

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, {}y begins to increase very quickly.

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Example Question #32 : Solving Exponential Functions

Solve: $2^{2x+3} = 8^{-6x}$

Possible Answers:

$-\frac{3}{20}$

$-\frac{3}{10}$

$\frac{11}{2}$

$-\frac{3}{5}$

$\frac{5}{3}$

Correct answer:

$-\frac{3}{20}$

Explanation:

The base of the right side can be rewritten as base two.

$2^{2x+3} = 2^{3(-6x)}$

2x+3 = 3(-6x)

Simplify the equation.

2x+3 = -18x

Add 18x on both sides.

2x+3 +18x= -18x+18x

20x+3 = 0

Subtract 3 on both sides.

20x+3 -3= 0-3

20x = -3

Divide by 20 on both sides.

$\frac{20x}{20} = \frac{-3}{20}$

The answer is: $-\frac{3}{20}$

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SAT II Math I : SAT Subject Test in Math I

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

All SAT II Math I Resources

Example Questions

Example Question #4 : Graphing Exponential Functions

Example Question #71 : Functions And Graphs

Example Question #6 : Graphing Exponential Functions

Example Question #361 : Sat Subject Test In Math I

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

Example Question #362 : Sat Subject Test In Math I

Example Question #32 : Solving Exponential Functions

All SAT II Math I Resources

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