Graphing an exponential function - GMAT Quantitative Reasoning

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Question

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

$y = 2^{x } - 20$

Answer

Set 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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Question

Define a function as follows:

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

Give the -intercept of the graph of .

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 .

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 .

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 .

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

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

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

Evaluate .

Answer

Rewrite the system as

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

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

and substitute and for $2 ^{x }$ and $3^{y}$ , respectively, to form the system

Add both sides:

$\underline{u- v= 43}$

.

Now backsolve:

Now substitute back:

$2 ^{x } = 16$

and

$3^{y}= 27$

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Question

Define a function as follows:

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

Give the -intercept of the graph of .

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 .

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.

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

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

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.

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

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

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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