GMAT Math : Graphing

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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)

The graph has no-intercept.

(10,0 )

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

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

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 :Graphing An Exponential Function

Define a functionas follows:

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

Give the-intercept of the graph of.

Possible Answers:

(2, 0)

(0,0)

The graph ofhas no-intercept.

(3,0)

$(\ln 3, 0)$

Correct answer:

The graph ofhas no-intercept.

Explanation:

Since the-intercept is the point at which the graph ofintersects 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 findsuch 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 oftherefore has no-intercept.

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Example Question #1 :How To Graph An Exponential Function

Define a functionas follows:

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

Give the vertical aysmptote of the graph of.

Possible Answers:

x= 3

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

x= 2

The graph ofdoes not have a vertical asymptote.

x = 5

Correct answer:

The graph ofdoes 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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Example Question #981 :Problem Solving Questions

Define a functionas follows:

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

Give the-intercept of the graph of.

Possible Answers:

(3,0)

(5,0)

(9,0)

The graph ofhas no-intercept.

(12,0)

Correct answer:

(5,0)

Explanation:

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

$3 ^{x-3}= 9$

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

x-3 = 2

x = 5

The-intercept is therefore (5,0) .

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Example Question #5 :How To Graph An Exponential Function

Define a functionas follows:

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

Give the horizontal aysmptote of the graph of.

Possible Answers:

y = 0

y = 1

y = 2

y = 7

$y = -\frac{7}{2}$

Correct answer:

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, $2 ^{x-1} > 0$ and $y= 2 ^{x-1} + 7 > 7$ for all real values of. The graph will never crosst the line of the equatin y = 7 , so this is the horizontal asymptote.

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Example Question #6 :How To Graph An Exponential Function

Define functionsandas follows:

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

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

Give the $y \,$ -coordinate of the point of intersection of their graphs.

Possible Answers:

Correct answer:

Explanation:

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

f(x) = g(x)

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

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

x+ 2= 2x-2

x+ 2 -x +2= 2x-2 -x +2

x = 4

To find the $y\,$ -coordinate, substitute 4 forin either definition:

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

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

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Example Question #133 :Exponential And Logarithmic Functions

Define a functionas follows:

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

Give the-intercept of the graph of.

Possible Answers:

(0,2)

(0,1)

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

(0,13)

(0,-3)

Correct answer:

(0,13)

Explanation:

The-coordinate ofthe-intercept of the graph ofis 0, and its-coordinate is f(0) :

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

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

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

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Example Question #7 :How To Graph An Exponential Function

Define functionsandas follows:

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

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

Give the $y \,$ -coordinate of the point of intersection of their graphs.

Possible Answers:

$3\sqrt{3}$

$9\sqrt{3}$

$\sqrt{3}$

Correct answer:

Explanation:

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

f(x) = g(x)

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

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

x+2 = -2x + 2

x+2 + 2x -2 = -2x + 2 + 2x -2

3x=0

x= 0

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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Example Question #1 :Graphing

Define a functionas follows:

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

Give the-intercept of the graph of.

Possible Answers:

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

(0,15)

$\left ( 0,125 \right )$

$\left ( 0,243 \right )$

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

Correct answer:

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

Explanation:

Since the-intercept is the point at which the graph ofintersects the-axis, the-coordinate is 0, and the-coordinate is f(0) :

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

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

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

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Example Question #9 :How To Graph An Exponential Function

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

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

Evaluate x+y .

Possible Answers:

The system has no solution.

Correct answer:

Explanation:

Rewrite the system as

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

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

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

2u + v= 59

u- v= -11

Add both sides:

2u + v= 5

$\underline{u- v= 43}$

=48

u = 16 .

Now backsolve:

16- v= -11

- v= - 27

v= 27

Now substitute back:

u = 16

$2 ^{x } = 16$

x = 4

and

v= 27

$3^{y}= 27$

y=3

x+y =4+3 = 7

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GMAT Math : Graphing

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 :Graphing

Example Question #1 :Graphing An Exponential Function

Example Question #1 :How To Graph An Exponential Function

Example Question #981 :Problem Solving Questions

Example Question #5 :How To Graph An Exponential Function

Example Question #6 :How To Graph An Exponential Function

Example Question #133 :Exponential And Logarithmic Functions

Example Question #7 :How To Graph An Exponential Function

Example Question #1 :Graphing

Example Question #9 :How To Graph An Exponential Function

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