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Advanced Geometry : Advanced Geometry

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6 Diagnostic Tests 57 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #111 : Advanced Geometry

$\small \small Which\ of\ the\ below\ represents\ the\ asymptote\ for\ y=log(x+4)-2$

Possible Answers:

$\small y=-2$

$\small x=4$

$\small x=-4$

$\small y=2$

Correct answer:

$\small x=-4$

Explanation:

$\small To\ find\ the\ asymptote\ for\ a\ log\ function\ we\ must\ know\ that\ the\ log\ of\ a\ negative\ number\ cannot\ exist.$

$\small So\ we\ set\ the\ inside\ of\ the\ log\ equal\ to\ zero\ and\ solve.$

$\small x+4=0$

$\small x=-4$

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Example Question #112 : Advanced Geometry

Give the range of the function

$f(x) = -3+ \log \left (x+ 2 \right )$ .

Possible Answers:

$\left \{ y | y \ge 2 \right \}$

$\left \{ y| y > 2 \right \}$

$\left \{ y | y > -3 \right \}$

The set of all real numbers

$\left \{ y | y \ge -3 \right \}$

Correct answer:

The set of all real numbers

Explanation:

Let $g(x) = \log x$

A logarithm can take any real value. Therefore, the range of g(x) is the set of real numbers, as is any linear transformation of g(x) .

In terms of g(x) ,

$f(x) = -3+ \log \left (x+ 2 \right ) = f(x) = g(x) -3$

making f(x) such a transformation. This makes the range of f(x) the set of all real numbers.

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Example Question #113 : Advanced Geometry

Give the domain of the equation

$f(x) = -3+ \log \left (x+ 2 \right )$ .

Possible Answers:

$\left \{ x | x \ge -3 \right \}$

$\left \{ x | x >- 2 \right \}$

$\left \{ x | x \ge -2 \right \}$

$\left \{ x | x > -3 \right \}$

The set of all real numbers

Correct answer:

$\left \{ x | x >- 2 \right \}$

Explanation:

A logarithm can be taken of a number if and only if the number is positive. Therefore,

$f(x) = -3+ \log \left (x+ 2 \right )$

is defined if and only if

x+ 2 > 0

or equivalently, subtracting 2 from both sides:

x+ 2 - 2 > 0 - 2

x >- 2

The correct domain is the set $\left \{ x | x >- 2 \right \}$ .

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Example Question #21 : How To Graph A Logarithm

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

$f(x) = -3+ \log \left (x+ 2 \right )$ .

Possible Answers:

x = -2

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

x = 0

x = -3

x= 1

Correct answer:

x = -2

Explanation:

Let $g(x) = \log x$

In terms of g(x) ,

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

This is the graph of g(x) shifted left 2 units ( ), then shifted down 3 units ( ).

The graph of g(x) has as its vertical asymptote the line of the equation x= 0 . The downward shift does not affect the position of this asymptote, but the left shift moves it down to the line of the equation x = -2 .

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Example Question #115 : Advanced Geometry

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

$f(x) = -3+ \log \left (x+ 2 \right )$ .

Possible Answers:

y = 0

y = 1

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

y = -2

y = -3

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) = g (x+ 2 ) - 3

This is the graph of g(x) shifted left 2 units, then shifted down 3 units.

The graph of g(x) does not have a horizontal asymptote; therefore, a transformation of this graph, such as that of f(x) , does not have a horizontal asymptote either.

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Example Question #26 : Graphing

Give the -coordinate of the -intercept of the graph of the equation

$f(x) = -3+ \log \left (x+ 2 \right )$ .

Possible Answers:

998

$\log \frac{1}{500}$

The graph of f(x) does not have a -intercept

$\log \frac{1}{998}$

Correct answer:

$\log \frac{1}{500}$

Explanation:

The -intercept of the graph of f(x) is the point at which it intersects the -axis. Its -coordinate is 0; its -coordinate is f(0) , which can be found by substituting 0 for in the definition:

$f(x) = -3+ \log \left (x+ 2 \right )$

$f(0) = -3+ \log \left (0+ 2 \right )$

$f(0) = -3+ \log 2$

$f(0) = \log 2 - 3$

This does not appear among the choices in this form. However, by definition,

$\log 10^{a} = a$ ,

so if

$\log 10^{a} = 3$ ,

a = 3 ,

and

$3 = \log 10^{3 } = \log 1,000$ .

And

$f(0) = \log 2 - \log 1,000$

By a property of logarithms, the difference of the logarithms of two numbers is equal to the logarithm of the quotient of those numbers, so

$f(0) = \log \frac{2}{1,000} = \log \frac{1}{500}$ .

This is the correct choice.

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Example Question #23 : Coordinate Geometry

Give the -coordinate of the -intercept of the graph of the equation

$f(x) = -3+ \log \left (x+ 2 \right )$ .

Possible Answers:

The graph of f(x) does not have an -intercept

$\log \frac{1}{500}$

998

Correct answer:

998

Explanation:

The -intercept(s) of the graph of f(x) are the point(s) at which it intersects the -axis. The -coordinate of each is 0; their -coordinate(s) are those value(s) of for which f(x) = 0 , so set up, and solve for , the equation:

f(x)= 0

$-3+ \log \left (x+ 2 \right ) = 0$

Add 3 to both sides:

$-3+ \log \left (x+ 2 \right ) + 3 = 0+ 3$

$\log \left (x+ 2 \right ) = 3$

"Log" indicates a common, or base ten, logarithm, so raise 10 to the power of both sides to eliminate the logarithm:

$10^{\log \left (x+ 2 \right ) }= 10 ^{3}$

x+2 = 1,000

Subtract 2 from both sides:

x+2 - 2 = 1,000 - 2

x = 998 ,

the correct choice.

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Example Question #1 : Graphing

Give the -intercept(s) of the graph of the equation

$y = 2^{x } - 20$

Possible Answers:

(-20,0)

(10,0 )

$\left ( 2 + \log_{2} 5, 0 \right )$

$\left ( 1 + \log_{2} 5, 0 \right )$

The graph has no -intercept.

Correct answer:

$\left ( 2 + \log_{2} 5, 0 \right )$

Explanation:

Set y = 0 and solve for :

$y = 2^{x } - 20$

$2^{x } - 20 = 0$

$2^{x } - 20+ 20 = 0 + 20$

$2^{x } = 20$

$\log_{2} \left ( 2 ^{x}\right ) = \log_{2} 20$

$x = \log_{2} 20$

$= \log_{2} 4 + \log_{2} 5$

$= 2 + \log_{2} 5$

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Example Question #1 : How To Graph An Exponential Function

Define a function as follows:

$f(x) = 3 ^{x-2}$

Give the -intercept of the graph of .

Possible Answers:

$(\ln 3, 0)$

The graph of has no -intercept.

(0,0)

(2, 0)

(3,0)

Correct answer:

The graph of has no -intercept.

Explanation:

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate can be found by setting f(x) equal to 0 and solving for . Therefore, we need to find such that $3 ^{x-2} = 0$ . However, any power of a positive number must be positive, so $3 ^{x-2} > 0$ for all real , and $3 ^{x-2} = 0$ has no real solution. The graph of therefore has no -intercept.

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Example Question #2 : Graphing

Define a function as follows:

$f(x) = 2 ^{x-3}+ 5$

Give the vertical aysmptote of the graph of .

Possible Answers:

The graph of does not have a vertical asymptote.

$x= - \frac{5}{2}$

x= 2

x = 5

x= 3

Correct answer:

The graph of does not have a vertical asymptote.

Explanation:

Since any number, positive or negative, can appear as an exponent, the domain of the function $f(x) = 2 ^{x-3}+ 5$ is the set of all real numbers; in other words, f(x) is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #111 : Advanced Geometry

Example Question #112 : Advanced Geometry

Example Question #113 : Advanced Geometry

Example Question #21 : How To Graph A Logarithm

Example Question #115 : Advanced Geometry

Example Question #26 : Graphing

Example Question #23 : Coordinate Geometry

Example Question #1 : Graphing

Example Question #1 : How To Graph An Exponential Function

Example Question #2 : Graphing

All Advanced Geometry Resources

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