Functions and Graphs - Algebra 2

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Question

Which of the following is true regarding the relation in the provided graph?

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Answer

A relation is a function if and only if it passes the Vertical Line Test (VLT)—that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

A function has an inverse, if and only if, it passes the Horizontal Line Test (HLT) - that is, no horizontal line exists that passes through its graph more than once. From the diagram below, we see at least one such line exists.

The function fails the HLT, so it does not have an inverse.

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Question

Which of the following is true of the graphed relationship?

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Answer

A relation is a function if and only if it passes the Vertical Line Test (VLT) - that is, no vertical line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

The relation passes the VLT, so it is a function.

A function has an inverse if and only if it passes the Horizontal Line Test (HLT) - that is, no horizontal line exists that passes through its graph more than once. From the diagram below, we see that no such line exists:

The function passes the HLT, so it has an inverse.

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Question

Above is the graph of a function . Which choice gives the graph of $f^{-1}(x)$ ?

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Answer

Given the graph of , the graph of its inverse, $f^{-1}(x)$ is the reflection of the former about the line . This line is in dark green below; critical points are reflected as shown:

The figure in blue is the graph of $f^{-1}(x)$

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Question

Above is the graph of a function . Which choice gives the graph of $f^{-1}(x)$ ?

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Answer

Given the graph of , the graph of its inverse, $f^{-1}(x)$ is the reflection of the former about the line . This line is in dark green below; critical points are reflected as shown:

The blue figure is $f^{-1}(x)$ , recreated below:

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Question

Define a function $f(x) = \sqrt{2x- 5}$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \sqrt{2x- 5}$

$y= \sqrt{2x- 5}$

Switch the positions of and :

$x= \sqrt{2y- 5}$ ,

or

$\sqrt{2y- 5} = x$

Solve for . This can be done as follows:

Square both sides:

$\left (\sqrt{2y- 5} \right ) ^{2} = x^{2}$

$2y- 5 = x^{2}$

Add 5 to both sides:

$2y- 5 + 5 = x^{2} + 5$

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

Multiply both sides by $\frac{1}{2}$ , distributing on the right:

$\frac{1}{2} \cdot 2y =\frac{1}{2} \cdot \left ( x^{2} + 5 \right )$

$y =\frac{1}{2} \cdot x^{2} + \frac{1}{2} \cdot 5$

$y =\frac{1}{2} x^{2} + \frac{5}{2}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =\frac{1}{2} x^{2} + \frac{5}{2}$ ,

the correct response.

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Question

Define a function $f(x) = 3 e ^{x} + 4$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = 3 e ^{x} + 4$

$y= 3 e ^{x} + 4$

Switch the positions of and :

$x= 3 e ^{y} + 4$

or

$3 e ^{y} + 4 = x$

Solve for - that is, isolate it on one side.

First, subtract 4:

$3 e ^{y} + 4 - 4 = x - 4$

$3 e ^{y} = x - 4$

Multiply by $\frac{1}{3}$ and distribute on the right:

$\frac{1}{3} \cdot 3 e ^{y} =\frac{1}{3} \cdot (x - 4)$

$e ^{y} =\frac{1}{3} \cdot x - \frac{1}{3} \cdot 4$

$e ^{y} =\frac{1}{3} x - \frac{4}{3}$

Take the natural logarithm of both sides:

$\ln e ^{y} =\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

$y =\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)=\ln \left (\frac{1}{3} x - \frac{4}{3} \right )$

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Question

Define a function $f(x) = e^{x-3}$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = e^{x-3}$

$y= e^{x-3}$

Switch the positions of and :

$x= e^{y-3}$ ,

or

$e^{y-3} = x$

Take the natural logarithm of both sides:

$\ln e^{y-3} = \ln x$

By definition, $\ln e^{a}= a$ , so

$y-3 = \ln x$

Add 3 to both sides:

$y-3 +3 = \ln x + 3$

$y = 3 + \ln x$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)= 3 + \ln x$

This is not given among the choices; however, remember that by one of the properties of logarithms,

$a = \ln e^{a}$ ,

so

$f^{-1}(x)= \ln e^{3} + \ln x$

By another property, $\ln M + \ln N = \ln MN$ , so

$f^{-1}(x)= \ln (e^{3} \cdot x)$

or

$f^{-1}(x)= \ln x e^{3}$ ,

which is among the choices and is the correct answer.

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Question

Define a function $f(x) = \frac{1}{5}x +7$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \frac{1}{5}x +7$

$y= \frac{1}{5}x +7$

Switch the positions of and :

$x= \frac{1}{5}y +7$

or,

$\frac{1}{5}y +7 = x$

Solve for - that is, isolate it on one side.

Subtract 7:

$\frac{1}{5}y +7 - 7 = x - 7$

$\frac{1}{5}y = x - 7$

Multiply by 5, distributing on the right:

$5 \cdot \frac{1}{5}y =5 \cdot (x - 7)$

$y =5 \cdot x - 5 \cdot 7$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =5x-35$

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Question

Define a function $f(x) = \frac{5}{x- 7}$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \frac{5}{x- 7}$

$y = \frac{5}{x- 7}$

Switch the positions of and :

$x = \frac{5}{y- 7}$ ,

or,

$\frac{5}{y- 7} = x$

Solve for - that is, isolate it on one side.

Take the reciprocals of both sides:

$\frac{y- 7} {5}= \frac{1}{x}$

Multiply both sides by 5:

$5 \cdot \frac{y- 7} {5}=5 \cdot \frac{1}{x}$

$y- 7 =\frac{5}{x}$

Add 7:

$y- 7 + 7 =\frac{5}{x} + 7$

$y =\frac{5}{x} + 7$

The right expression can be simplified as follows:

$y =\frac{5}{x} + \frac{7x}{x}$

$y =\frac{7x+ 5 }{x}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =\frac{7x+ 5 }{x}$

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Question

Define a function $f(x) = 5^{x} + 5$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = 5^{x} + 5$

$y = 5^{x} + 5$

Switch the positions of and :

$x = 5^{y} + 5$ ,

or

$5^{y} + 5 = x$

Solve for - that is, isolate it on one side - as follows:

First, subtract 5 from both sides:

$5^{y} + 5- 5 = x - 5$

$5^{y} = x - 5$

Take the base-5 logarithm of both sides:

$\log_{5} 5^{y} = \log_{5} (x - 5)$

A property of logarithms states that $\log_{a} a^{N}= N$ , so

$y = \log_{5} (x - 5)$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) = \log_{5} (x - 5)$

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Question

Define a function $f(x) = \ln (3x+ 7)$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \ln (3x+ 7)$

$y= \ln (3x+ 7)$

Switch the positions of and :

$x= \ln (3y+ 7)$

or

$\ln (3y+ 7) = x$

Solve for - that is, isolate it on one side - as follows:

Raise to the power of both sides:

$e^{ \ln (3y+ 7)} = e^{x}$

A property of logarithms states that $e ^{\log N} = N$ , so

$3y+ 7 = e^{x}$

Subtract 7 from both sides:

$3y+ 7 -7 = e^{x} - 7$

$3y = e^{x} - 7$

Multiply both sides by $\frac{1}{3}$ , distributing on the right:

$\frac{1}{3} \cdot 3y =\frac{1}{3} \cdot( e^{x} - 7)$

$y =\frac{1}{3} \cdot e^{x} - \frac{1}{3} \cdot7$

$y =\frac{1}{3} e^{x} - \frac{7}{3}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =\frac{1}{3} e^{x} - \frac{7}{3}$

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Question

Define a function $f(x) = \frac{x-2}{2x}$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \frac{x-2}{2x}$

$y= \frac{x-2}{2x}$

Switch the positions of and :

$x= \frac{y-2}{2y}$

Solve for - that is, isolate it on one side - as follows:

Split the expression at right into the difference of two separate expressions:

$x = \frac{y}{2y} - \frac{2}{2y}$

Simplify:

$x = \frac{1}{2} - \frac{1}{y}$

Add $\frac{1}{y} - x$ to both sides:

$x + \frac{1}{y} - x = \frac{1}{2} - \frac{1}{y} + \frac{1}{y} - x$

$\frac{1}{y} = \frac{1}{2} - x$

Simplify the expression at right:

$\frac{1}{y} = \frac{1}{2} - \frac{2x}{2}$

$\frac{1}{y} = \frac{1-2x}{2}$

Take the reciprocal of both sides:

$y= \frac{2}{1-2x}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) = \frac{2}{1-2x}$

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Question

Define a function $f(x) = \sqrt[3]{4x-9}$ .

Which statement correctly gives $f^{-1}(x)$ ?

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \sqrt[3]{4x-9}$

$y = \sqrt[3]{4x-9}$

Switch the positions of and :

$x = \sqrt[3]{4y-9}$

or

$\sqrt[3]{4y-9} = x$

Solve for - that is, isolate it on one side - as follows:

Raise both sides to the third power:

$\left (\sqrt[3]{4y-9} \right )^{3}= x^{3}$

$4y-9 = x^{3}$

Add 9 to both sides:

$4y-9 + 9 = x^{3} + 9$

$4y = x^{3} + 9$

Multiply both sides by $\frac{1}{4}$ , distributing on the right side:

$\frac{1}{4} \cdot 4y =\frac{1}{4} \cdot ( x^{3} + 9)$

$y =\frac{1}{4} \cdot x^{3} + \frac{1}{4} \cdot 9$

$y =\frac{1}{4} x^{3} + \frac{9}{4}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =\frac{1}{4} x^{3} + \frac{9}{4}$

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Question

Define a function $f(x)= \frac{x}{x-1}$ .

True or false: is its own inverse.

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Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x)= \frac{x}{x-1}$

$y = \frac{x}{x-1}$

Switch and :

$x = \frac{y}{y-1}$

Solve for - that is, isolate on one side of the equation - as follows:

Multiply both sides by , distributing on the right side:

$x\cdot (y-1) = \frac{y}{y-1} \cdot (y-1)$

$x\cdot y-x\cdot 1 = y$

Add to both sides to get all terms to the left, then factor out :

$x \cdot y -1 \cdot y =x$

Divide both sides by :

$\frac{(x -1 ) y}{x -1 } =\frac{x}{x -1 }$

$y =\frac{x}{x -1 }$

Replace with $f^{-1}(x)$ :

$f^{-1}(x) =\frac{x}{x -1 }$

Therefore, $f^{-1}(x)= f(x)$ , and is indeed its own inverse.

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Question

What is the equation of the line displayed above?

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Answer

The equation of a line is , with m being the slope of the line, and b being the y-intercept. The y intercept of the line is at , so .

The x-intercept is at , the equation becomes , simplification yields $m=\frac{1}{3}$

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Question

$f(x)=x^{2}$

$g(x)=4x^{2}-3$

How is the graph of different from the graph of ?

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Answer

Almost all transformed functions can be written like this:

where is the parent function. In this case, our parent function is $f(x)=x^{2}$ , so we can write this way:

$g(x)=a[b(x-c)]^{2}+d$

Luckily, for this problem, we only have to worry about and .

represents the vertical stretch factor of the graph.

If is less than 1, the graph has been vertically compressed by a factor of . It's almost as if someone squished the graph while their hands were on the top and bottom. This would make a parabola, for example, look wider.

If is greater than 1, the graph has been vertically stretched by a factor of . It's almost as if someone pulled on the graph while their hands were grasping the top and bottom. This would make a parabola, for example, look narrower.

represents the vertical translation of the graph.

If is positive, the graph has been shifted up units.

If is negative, the graph has been shifted down units.

For this problem, is 4 and is -3, causing vertical stretch by a factor of 4 and a vertical translation down 3 units.

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Question

Which of the following represents a standard parabola shifted up by 2 units?

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Answer

Begin with the standard equation for a parabola: .

Vertical shifts to this standard equation are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift upward 2 units, add 2.

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Question

Which of the following transformation flips a parabola vertically, doubles its width, and shifts it up by 3?

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Answer

Begin with the standard equation for a parabola: .

Inverting, or flipping, a parabola refers to the sign in front of the coefficient of the . If the coefficient is negative, then the parabola opens downward.

The width of the parabola is determined by the magnitude of the coefficient in front of . To make a parabola wider, use a fractional coefficient. Doubling the width indicates a coefficient of one-half.

$f(x)=-(\frac{1}{2})x^2=-0.5x^2$

Vertical shifts are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift upward 3 units, add 3.

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Question

Which of the following shifts a parabola six units to the right and five downward?

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Answer

Begin with the standard equation for a parabola: .

Vertical shifts to this standard equation are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift downward 5 units, subtract 5.

Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the term. To shift 6 units to the right, subtract 6 within the parenthesis.

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Question

Which of the following transformations represents a parabola shifted to the right by 4 and halved in width?

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Answer

Begin with the standard equation for a parabola: .

Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the term. To shift 4 units to the right, subtract 4 within the parenthesis.

The width of the parabola is determined by the magnitude of the coefficient in front of . To make a parabola narrower, use a whole number coefficient. Halving the width indicates a coefficient of two.

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