GMAT Math : Graphing an exponential function

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Example Questions

Example Question #1 : Graphing

Give the \(\displaystyle x\)-intercept(s) of the graph of the equation 

\(\displaystyle y = 2^{x } - 20\)

Possible Answers:

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

\(\displaystyle (-20,0)\)

The graph has no \(\displaystyle x\)-intercept.

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

\(\displaystyle (10,0 )\)

Correct answer:

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

Explanation:

Set \(\displaystyle y = 0\) and solve for \(\displaystyle x\):

\(\displaystyle y = 2^{x } - 20\)

\(\displaystyle 2^{x } - 20 = 0\)

\(\displaystyle 2^{x } - 20+ 20 = 0 + 20\)

\(\displaystyle 2^{x } = 20\)

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

\(\displaystyle x = \log_{2} 20\)

\(\displaystyle = \log_{2} 4 + \log_{2} 5\)

\(\displaystyle = 2 + \log_{2} 5\)

Example Question #1 : How To Graph An Exponential Function

Define a function \(\displaystyle f\) as follows:

\(\displaystyle f(x) = 3 ^{x-2}\)

Give the \(\displaystyle x\)-intercept of the graph of \(\displaystyle f\).

Possible Answers:

\(\displaystyle (3,0)\)

\(\displaystyle (\ln 3, 0)\)

\(\displaystyle (0,0)\)

\(\displaystyle (2, 0)\)

The graph of \(\displaystyle f\) has no \(\displaystyle x\)-intercept. 

Correct answer:

The graph of \(\displaystyle f\) has no \(\displaystyle x\)-intercept. 

Explanation:

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

Example Question #2 : Graphing

Define a function \(\displaystyle f\) as follows:

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

Give the vertical aysmptote of the graph of \(\displaystyle f\).

Possible Answers:

\(\displaystyle x= 3\)

\(\displaystyle x= 2\)

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

\(\displaystyle x= - \frac{5}{2}\)

\(\displaystyle x = 5\)

Correct answer:

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

Explanation:

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

Example Question #1 : How To Graph An Exponential Function

Define a function \(\displaystyle f\) as follows:

\(\displaystyle f(x) = 3 ^{x-3}- 9\)

Give the \(\displaystyle x\)-intercept of the graph of \(\displaystyle f\).

Possible Answers:

\(\displaystyle (12,0)\)

\(\displaystyle (5,0)\)

\(\displaystyle (3,0)\)

\(\displaystyle (9,0)\)

The graph of \(\displaystyle f\) has no \(\displaystyle x\)-intercept. 

Correct answer:

\(\displaystyle (5,0)\)

Explanation:

Since the \(\displaystyle x\)-intercept is the point at which the graph of \(\displaystyle f\) intersects the \(\displaystyle x\)-axis, the \(\displaystyle y\)-coordinate is 0, and the \(\displaystyle x\)-coordinate can be found by setting \(\displaystyle f(x)\) equal to 0 and solving for \(\displaystyle y\). Therefore, we need to find \(\displaystyle x\) such that 

\(\displaystyle 3 ^{x-3}- 9 = 0\)

\(\displaystyle 3 ^{x-3}= 9\)

\(\displaystyle 3 ^{x-3}= 3 ^{2}\)

\(\displaystyle x-3 = 2\)

\(\displaystyle x = 5\)

The \(\displaystyle x\)-intercept is therefore \(\displaystyle (5,0)\).

Example Question #5 : Graphing An Exponential Function

Define a function \(\displaystyle f\) as follows:

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

Give the horizontal aysmptote of the graph of \(\displaystyle f\).

Possible Answers:

\(\displaystyle y = 1\)

\(\displaystyle y = 7\)

\(\displaystyle y = 2\)

\(\displaystyle y = 0\)

\(\displaystyle y = -\frac{7}{2}\)

Correct answer:

\(\displaystyle y = 7\)

Explanation:

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

Example Question #6 : Graphing An Exponential Function

Define functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle f(x) = 2^{x +2}\)

\(\displaystyle g(x) = 4^{x-1}\)

Give the \(\displaystyle y \,\)-coordinate of the point of intersection of their graphs.

Possible Answers:

\(\displaystyle 8\)

\(\displaystyle 16\)

\(\displaystyle 4\)

\(\displaystyle 64\)

\(\displaystyle 32\)

Correct answer:

\(\displaystyle 64\)

Explanation:

First, we rewrite both functions with a common base:

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

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

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

\(\displaystyle g(x) = 2 ^{2 (x-1)}\)

\(\displaystyle g(x) = 2 ^{2x-2}\)

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

\(\displaystyle f(x) = g(x)\)

\(\displaystyle 2^{x +2}= 2 ^{2x-2}\)

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

\(\displaystyle x+ 2= 2x-2\)

\(\displaystyle x+ 2 -x +2= 2x-2 -x +2\)

\(\displaystyle x = 4\)

To find the \(\displaystyle y\,\)-coordinate, substitute 4 for \(\displaystyle x\) in either definition:

\(\displaystyle f(x) = 2^{x +2}\)

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

Example Question #7 : Graphing An Exponential Function

Define a function \(\displaystyle f\) as follows:

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

Give the \(\displaystyle y\)-intercept of the graph of \(\displaystyle f\).

Possible Answers:

\(\displaystyle (0,13)\)

\(\displaystyle (0,2)\)

\(\displaystyle (0,1)\)

\(\displaystyle (0,-3)\)

\(\displaystyle \left ( 0, \frac{3}{4}\right )\)

Correct answer:

\(\displaystyle (0,13)\)

Explanation:

The \(\displaystyle x\)-coordinate ofthe \(\displaystyle y\)-intercept of the graph of \(\displaystyle f\) is 0, and its \(\displaystyle y\)-coordinate is \(\displaystyle f(0)\):

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

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

The \(\displaystyle y\)-intercept is the point \(\displaystyle (0,13)\).

Example Question #2 : How To Graph An Exponential Function

Define functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle f(x) = 3^{x +2}\)

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

Give the \(\displaystyle y \,\)-coordinate of the point of intersection of their graphs.

Possible Answers:

\(\displaystyle 3\sqrt{3}\)

\(\displaystyle 9\sqrt{3}\)

\(\displaystyle 3\)

\(\displaystyle \sqrt{3}\)

\(\displaystyle 9\)

Correct answer:

\(\displaystyle 9\)

Explanation:

First, we rewrite both functions with a common base:

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

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

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

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

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

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

\(\displaystyle f(x) = g(x)\)

\(\displaystyle 3^{x +2} = 3^{-2 x+2}\)

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

\(\displaystyle x+2 = -2x + 2\)

\(\displaystyle x+2 + 2x -2 = -2x + 2 + 2x -2\)

\(\displaystyle 3x=0\)

\(\displaystyle x= 0\)

Now substitute in either function:

\(\displaystyle f(x) = 3^{x +2}\)

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

 

Example Question #9 : Graphing An Exponential Function

Define a function \(\displaystyle f\) as follows:

\(\displaystyle f(x) =5^{x-3}\)

Give the \(\displaystyle y\)-intercept of the graph of \(\displaystyle f\).

Possible Answers:

\(\displaystyle \left ( 0, \frac{1}{243} \right )\)

\(\displaystyle \left ( 0,125 \right )\)

\(\displaystyle (0,15)\)

\(\displaystyle \left ( 0, \frac{1}{125} \right )\)

\(\displaystyle \left ( 0,243 \right )\)

Correct answer:

\(\displaystyle \left ( 0, \frac{1}{125} \right )\)

Explanation:

Since the \(\displaystyle y\)-intercept is the point at which the graph of \(\displaystyle f\) intersects the \(\displaystyle y\)-axis, the \(\displaystyle x\)-coordinate is 0, and the \(\displaystyle y\)-coordinate is \(\displaystyle f(0)\):

\(\displaystyle f(x) =5^{x-3}\)

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

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

Example Question #10 : Graphing An Exponential Function

\(\displaystyle 2 ^{x+1} + 3^{y} = 59\)

\(\displaystyle 2 ^{x } - 3^{y} = -11\)

Evaluate \(\displaystyle x+y\).

Possible Answers:

\(\displaystyle 43\)

\(\displaystyle 11\)

The system has no solution.

\(\displaystyle 7\)

\(\displaystyle 12\)

Correct answer:

\(\displaystyle 7\)

Explanation:

Rewrite the system as 

\(\displaystyle 2 \cdot 2 ^{x} + 3^{y} = 59\)

\(\displaystyle 2 ^{x } - 3^{y} = -11\)

and substitute \(\displaystyle u\) and \(\displaystyle v\) for \(\displaystyle 2 ^{x }\) and \(\displaystyle 3^{y}\), respectively, to form the system

\(\displaystyle 2u + v= 59\)

\(\displaystyle u- v= -11\)

Add both sides:

\(\displaystyle 2u + v= 5\)

\(\displaystyle \underline{u- v= 43}\)

\(\displaystyle 3u\)        \(\displaystyle =48\)

\(\displaystyle u = 16\).

Now backsolve:

\(\displaystyle 16- v= -11\)

\(\displaystyle - v= - 27\)

\(\displaystyle v= 27\)

Now substitute back:

\(\displaystyle u = 16\)

\(\displaystyle 2 ^{x } = 16\)

\(\displaystyle x = 4\)

and

\(\displaystyle v= 27\)

\(\displaystyle 3^{y}= 27\)

\(\displaystyle y=3\)

\(\displaystyle x+y =4+3 = 7\)

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