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College Algebra : College Algebra

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Example Questions

Example Question #4 : Transformations

Consider an exponential function y=e^x . If we want to reflect this function across the y-axis, which of the following equations would result in the desired reflection?

Possible Answers:

$y=-e^{-x}$

$y=e^{-x}$

y=-e^x

y=e^x -1

y=e^x +1

Correct answer:

$y=e^{-x}$

Explanation:

As a general rule, if you have a function f(x) , then in order to reflect across the x-axis, we compute -f(x) , and in order to reflect across the y-axis, we compute f(-x) . In our case, we are asked to compute the latter.

So, if f(x)=e^x , then $f(-x)=e^{-x}$ .

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Example Question #5 : Transformations

If we want a function y=(x-1)^2 +2 to be reflected about the origin, what would the corresponding equation look like?

Possible Answers:

y=-[(x-1)^2 -2]

y=(-x-1)^2 +2

y=-[(-x-1)^2 +2]

y=-(x-1)^2 -2

y=[(x-1)^2 -2]

Correct answer:

y=-[(-x-1)^2 +2]

Explanation:

To compute a reflection about the x-axis, calculate -f(x) , and to calculate a reflection about the y-axis, calculate f(-x) . To compute a reflection about the origin, simply combine both reflections into -f(-x) .

In our case, f(x)=(x-1)^2 +2 .

So, -f(-x)=-[(-x-1)^2 +2]

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Example Question #61 : Graphs

Give the equation of the horizontal asymptote of the graph of the equation

$f(x) =4 \log (x- 3)- 2$ .

Possible Answers:

The graph of f(x) does not have a horizontal asymptote.

y = 10

y = 3

y = 2

y = 4

Correct answer:

The graph of f(x) does not have a horizontal asymptote.

Explanation:

Let $g(x) = \log x$ . In terms of g(x) ,

f(x) =4 g (x- 3)- 2

g(x) , being a logarithmic function, has a graph without a horizontal asymptote. As f(x) represents the result of transformations of g(x) , it follows that its graph does not have a horizontal asymptote, either.

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Example Question #61 : Graphs

Give the equation of the vertical asymptote of the graph of the equation

$f(x) =4 \log (x- 3)- 2$ .

Possible Answers:

x = 4

x = 2

x = 3

x = -3

x = -2

Correct answer:

x = 3

Explanation:

Let $g(x) = \log x$ . In terms of g(x) ,

f(x) =4 g (x- 3)- 2

The graph of g(x) has as its vertical asymptote the line of the equation x= 0 . The graph of f(x) is the result of three transformations on the graph of g(x) - a right shift of 3 units ( ), a vertical stretch ( ), and a downward shift of 2 units ( ). Of the three transformations, only the right shift affects the position of the vertical asymptote; the asymptote of f(x) also shifts right 3 units, to x = 3 .

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Example Question #1 : Transformations

Define

$f(x) = e^{x}$

and

$g(x)= -e^{x-2}$ .

Which two transformations must be performed in the graph of f(x) in order to obtain the graph of g(x) ?

Possible Answers:

None of the other choices gives the correct response.

The graph of f(x) must be translated two units right, then reflected about the -axis.

The graph of f(x) must be translated two units left, then reflected about the -axis.

Correct answer:

The graph of f(x) must be translated two units right, then reflected about the -axis.

Explanation:

g(x) = -f(x-2) , so the graph of g(x) is the result of performing the following transformations:

1) $f_{1}(x) = f(x-2)$ is the result of translating this graph two units right.

2) $g(x)= -f_{1}(x) = -f(x-2)$ is the result of reflecting the new graph about the -axis.

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Example Question #11 : Transformations

The graph of a function f(x) is reflected about the -axis, then translated upward $\frac{8}{3}$ units. Which of the following is represented by the resulting graph?

Possible Answers:

$g(x)= -f(x) + \frac{8}{3}$

$g(x)= -f\left (x- \frac{8}{3} \right )$

$g(x)= -f\left (-x + \frac{8}{3} \right )$

$g(x)= f(-x) + \frac{8}{3}$

$g(x)= -f\left (x + \frac{8}{3} \right )$

Correct answer:

$g(x)= -f(x) + \frac{8}{3}$

Explanation:

Reflecting the graph of a function f(x) about the -axis results in the graph of the function

$f_{1}(x)= -f(x)$ .

Translating this graph upward $\frac{8}{3}$ results in the graph of the function

$g(x)= f_{1}(x) + \frac{8}{3 }= -f(x)+ \frac{8}{3 }$ .

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Example Question #63 : Graphs

Translate the graph of $f(x) = \ln x$ upward three units to yield the graph of a function g(x) . Which of the following is a valid way of stating the definition of g(x) ?

Possible Answers:

$g(x)= \ln (x+3)$

$g(x)= \ln (e^{3} x)$

$g(x)= \ln \left ( \frac{x}{e^{3}} \right )$

$g(x)= \ln (x-3)$

$g(x)= \ln x^{3}$

Correct answer:

$g(x)= \ln (e^{3} x)$

Explanation:

A vertical translation of the graph of a function f(x) by units yields the graph of the function g(x)+k . A translation in an upward direction is a positive translation, so setting $f(x)= \ln x$ and k = 3 , the resulting graph becomes

$g(x)=( \ln x) +3$

$g(x)=3+ \ln x$

Apply properties of logarithms to rewrite this as

$g(x)= \ln e^{3} + \ln x$

$g(x)= \ln (e^{3} x)$ .

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Example Question #11 : Transformations

Reflect the graph of $f(x) = \ln x$ about the -axis to yield the graph of a function g(x) . Which of the following is a valid way of stating the definition of g(x) ?

Possible Answers:

$g(x) =e^{x}$

$g(x)= \ln (-x)$

None of the other choices gives the correct response.

$g(x) = \frac{1}{\ln x}$

$g(x)= \ln \frac{1}{x}$

Correct answer:

$g(x)= \ln (-x)$

Explanation:

The reflection of the graph of a function f(x) about the -axis yields the graph of the function g(x) = f(-x) . Therefore, set $f(x) = \ln x$ and substitute for to yield the function

$g(x) = \ln (-x)$ .

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Example Question #1 : Symmetry

Determine the symmetry of the following equation.

y=x^2-6x+10

Possible Answers:

Symmetry along the x-axis.

Does not have symmetry.

Symmetry along the origin.

Symmetry along all axes.

Symmetry along the y-axis.

Correct answer:

Does not have symmetry.

Explanation:

To check for symmetry, we are going to do three tests, which involve substitution. First one will be to check symmetry along the x-axis, replace -y=y .

-y=x^2-6x+10

This isn't equivilant to the first equation, so it's not symmetric along the x-axis.

Next is to substitute -x=x .

y=(-x)^2-6(-x)+10

y=x^2+6x+10

This is not the same, so it is not symmetric along the y-axis.

For the last test we will substitute -y=y , and -x=x

-y=(-x)^2-6(-x)+10

-y=x^2+6x+10

y=-x^2-6x-10

This isn't the same as the orginal equation, so it is not symmetric along the origin.

The answer is it is not symmetric along any axis.

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Example Question #1 : Symmetry

Untitled

Which of the following is true of the relation graphed above?

Possible Answers:

It is an odd function

It is not a function

It is an even function

It is a function, but it is neither even nor odd.

Correct answer:

It is an odd function

Explanation:

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Untitled

Also, it can be seen to be symmetrical about the origin. Consequently, for each in the domain, f(-c) = -f(c) - the function is odd.

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College Algebra : College Algebra

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #4 : Transformations

Example Question #5 : Transformations

Example Question #61 : Graphs

Example Question #61 : Graphs

Example Question #1 : Transformations

Example Question #11 : Transformations

Example Question #63 : Graphs

Example Question #11 : Transformations

Example Question #1 : Symmetry

Example Question #1 : Symmetry

All College Algebra Resources

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