Graphing an exponential function - GMAT Quantitative

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Question

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

$y = $2^{x }$ - 20$

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Answer

Set and solve for :

$y = $2^{x }$ - 20$

$x = $\log_{2}$ 20$

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

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

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

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Answer

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate is :

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

$f(0) $=5^{0-3}$= $5^{-3}$ = $$\frac{1}{5^{3}$$} = $\frac{1}{125}$$ ,

The -intercept is the point $\left ( 0, $\frac{1}{125}$ \right )$ .

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

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Answer

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 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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Question

Define a function as follows:

$f(x) = 2 ^{x-1} + 7$

Give the horizontal aysmptote of the graph of .

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Answer

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, $2 ^{x-1} > 0$ and $y= 2 ^{x-1} + 7 > 7$ for all real values of . The graph will never crosst the line of the equatin , so this is the horizontal asymptote.

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Question

Define a function as follows:

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

Give the vertical aysmptote of the graph of .

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Answer

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, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

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Answer

The -coordinate ofthe -intercept of the graph of is 0, and its -coordinate is :

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

$f(0) = 4 ^{0+2} - 3 = 4 ^{2} - 3 = 16 - 3 = 13$

The -intercept is the point .

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Question

Define a function as follows:

$f(x) = 3 ^{x-3}- 9$

Give the -intercept of the graph of .

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Answer

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 equal to 0 and solving for . Therefore, we need to find such that

$3 ^{x-3}- 9 = 0$ .

$3 ^{x-3}= 9$

$3 ^{x-3}= 3 ^{2}$

The -intercept is therefore .

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Question

Define functions and as follows:

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

$g(x) = $4^{x-1}$$

Give the -coordinate of the point of intersection of their graphs.

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Answer

First, we rewrite both functions with a common base:

$f(x) = $2^{x +2}$$ is left as it is.

$g(x) = $4^{x-1}$$ can be rewritten as

$g(x) = \left (2 ^{2} \right $)^{x-1}$$

$g(x) = 2 ^{2 (x-1)}$

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

To find the point of intersection of the graphs of the functions, set

The powers are equal and the bases are equal, so we can set the exponents equal to each other and solve:

To find the -coordinate, substitute 4 for in either definition:

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

$f(4) = $2^{4+2}$= $2^{6}$ = 64$ , the correct response.

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Question

Define functions and as follows:

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

$g(x) =\left ( $\frac{1}{9}$ \right $)^{x-1}$$

Give the -coordinate of the point of intersection of their graphs.

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Answer

First, we rewrite both functions with a common base:

$f(x) = $3^{x +2}$$ is left as it is.

$g(x) =\left ( $\frac{1}{9}$ \right $)^{x-1}$$ can be rewritten as

$g(x) =\left ( $3^{-2}$ \right $)^{x-1}$$

$g(x) = $3^{-2 (x-1)}$$

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

To find the point of intersection of the graphs of the functions, set

Since the powers of the same base are equal, we can set the exponents equal:

Now substitute in either function:

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

$f(0) = $3^{0 +2}$ = $3^{ 2}$ = 9$ , the correct answer.

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Question

$2 ^{x+1} + $3^{y}$ = 59$

$2 ^{x } - $3^{y}$ = -11$

Evaluate .

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Answer

Rewrite the system as

$2 \cdot 2 ^{x} + $3^{y}$ = 59$

$2 ^{x } - $3^{y}$ = -11$

and substitute and for $2 ^{x }$ and , respectively, to form the system

Add both sides:

$\underline{u- v= 43}$

.

Now backsolve:

Now substitute back:

$2 ^{x } = 16$

and

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Question

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

$y = $2^{x }$ - 20$

Tap to reveal answer

Answer

Set and solve for :

$y = $2^{x }$ - 20$

$x = $\log_{2}$ 20$

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

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

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to reveal answer

Answer

Since the -intercept is the point at which the graph of intersects the -axis, the -coordinate is 0, and the -coordinate is :

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

$f(0) $=5^{0-3}$= $5^{-3}$ = $$\frac{1}{5^{3}$$} = $\frac{1}{125}$$ ,

The -intercept is the point $\left ( 0, $\frac{1}{125}$ \right )$ .

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to reveal answer

Answer

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 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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Question

Define a function as follows:

$f(x) = 2 ^{x-1} + 7$

Give the horizontal aysmptote of the graph of .

Tap to reveal answer

Answer

The horizontal asymptote of an exponential function can be found by noting that a positive number raised to any power must be positive. Therefore, $2 ^{x-1} > 0$ and $y= 2 ^{x-1} + 7 > 7$ for all real values of . The graph will never crosst the line of the equatin , so this is the horizontal asymptote.

← Didn't Know|Knew It →

Question

Define a function as follows:

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

Give the vertical aysmptote of the graph of .

Tap to reveal answer

Answer

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, is defined for all real values of . It is therefore impossible for the graph to have a vertical asymptote.

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to reveal answer

Answer

The -coordinate ofthe -intercept of the graph of is 0, and its -coordinate is :

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

$f(0) = 4 ^{0+2} - 3 = 4 ^{2} - 3 = 16 - 3 = 13$

The -intercept is the point .

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Question

Define a function as follows:

$f(x) = 3 ^{x-3}- 9$

Give the -intercept of the graph of .

Tap to reveal answer

Answer

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 equal to 0 and solving for . Therefore, we need to find such that

$3 ^{x-3}- 9 = 0$ .

$3 ^{x-3}= 9$

$3 ^{x-3}= 3 ^{2}$

The -intercept is therefore .

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Question

Define functions and as follows:

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

$g(x) = $4^{x-1}$$

Give the -coordinate of the point of intersection of their graphs.

Tap to reveal answer

Answer

First, we rewrite both functions with a common base:

$f(x) = $2^{x +2}$$ is left as it is.

$g(x) = $4^{x-1}$$ can be rewritten as

$g(x) = \left (2 ^{2} \right $)^{x-1}$$

$g(x) = 2 ^{2 (x-1)}$

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

To find the point of intersection of the graphs of the functions, set

The powers are equal and the bases are equal, so we can set the exponents equal to each other and solve:

To find the -coordinate, substitute 4 for in either definition:

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

$f(4) = $2^{4+2}$= $2^{6}$ = 64$ , the correct response.

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Question

Define functions and as follows:

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

$g(x) =\left ( $\frac{1}{9}$ \right $)^{x-1}$$

Give the -coordinate of the point of intersection of their graphs.

Tap to reveal answer

Answer

First, we rewrite both functions with a common base:

$f(x) = $3^{x +2}$$ is left as it is.

$g(x) =\left ( $\frac{1}{9}$ \right $)^{x-1}$$ can be rewritten as

$g(x) =\left ( $3^{-2}$ \right $)^{x-1}$$

$g(x) = $3^{-2 (x-1)}$$

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

To find the point of intersection of the graphs of the functions, set

Since the powers of the same base are equal, we can set the exponents equal:

Now substitute in either function:

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

$f(0) = $3^{0 +2}$ = $3^{ 2}$ = 9$ , the correct answer.

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Question

$2 ^{x+1} + $3^{y}$ = 59$

$2 ^{x } - $3^{y}$ = -11$

Evaluate .

Tap to reveal answer

Answer

Rewrite the system as

$2 \cdot 2 ^{x} + $3^{y}$ = 59$

$2 ^{x } - $3^{y}$ = -11$

and substitute and for $2 ^{x }$ and , respectively, to form the system

Add both sides:

$\underline{u- v= 43}$

.

Now backsolve:

Now substitute back:

$2 ^{x } = 16$

and

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