How to graph an exponential function - Geometry

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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

If the functions

$f(x) = e^{x}+ 9$

$g(x) = 4e^{x}- 7$

were graphed on the same coordinate axes, what would be the -coordinate of their point of intersection?

Round to the nearest tenth, if applicable.

Answer

We can rewrite the statements using for both and as follows:

$y = e^{x}+ 9$

$y = 4e^{x}- 7$

To solve this, we can set the expressions equal, as follows:

$4e^{x}- 7 = e^{x}+ 9$

$4e^{x}- 7 - e^{x} + 7 = e^{x}+ 9 - e^{x} + 7$

$3e^{x} = 16$

$3e^{x}\div 3 = 16 \div 3$

$e^{x} = \approc 5.3333$

$x \approx \ln 5.333 \approx 1.7$

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Question

If the functions

$f(x) = e^{x}+ 9$

$g(x)= 4e^{x}- 7$

were graphed on the same coordinate axes, what would be the -coordinate of their point of intersection?

Round to the nearest tenth, if applicable.

Answer

We can rewrite the statements using for both and as follows:

$y = e^{x}+ 9$

$y = 4e^{x}- 7$

To solve this, we can multiply the first equation by , then add:

$-4y = -4e^{x}-36$

$\underline{y = 4e^{x} ; ; - 7}$

$-3y \div (-3) = -43 \div (-3 )$

$y \approx 14.3$

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Question

Give the -intercept of the graph of the function

$f(x) = 5 \cdot 4^{x- 3}- 3$

Round to the nearest tenth, if applicable.

Answer

The -intercept is , where :

$f(x) = 5 \cdot 4^{x- 3}- 3$

$5 \cdot 4^{a- 3}- 3 = 0$

$5 \cdot 4^{a- 3}- 3 + 3= 0 + 3$

$5 \cdot 4^{a- 3}= 3$

$5 \cdot 4^{a- 3} \div 5= 3 \div 5$

$4^{a- 3} = 0.6$

$\ln \left (4^{a- 3} \right )=\ln 0.6$

$(a- 3)\ln 4 =\ln 0.6$

$a- 3=\frac{\ln 0.6}{\ln 4 } \approx \frac{-0.5108}{1.3863}\approx -0.3685$

$a- 3+3 \approx -0.3685 + 3$

$a \approx 2.63$

The -intercept is .

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Question

Give the -intercept of the graph of the function

$f(x) = 5 \cdot 4^{x- 3}- 3$

Round to the nearest hundredth, if applicable.

Answer

The -intercept is :

$f(x) = 5 \cdot 4^{x- 3}- 3$

$f(0) = 5 \cdot 4^{0- 3}- 3$

$f(0) = 5 \cdot 4^{ - 3}- 3$

$f(0) = 5 \cdot \frac{1}{64}- 3$

$f(0) = \frac{5}{64}- 3$

$f(0) \approx 0.08 - 3 \approx -2.92$

is the -intercept.

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Question

Give the horizontal asymptote of the graph of the function

$g(x) = 9 \cdot 3^{x}- 3$

Answer

We can rewrite this as follows:

$g(x) = 9 \cdot 3^{x}- 3$

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

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

This is a translation of the graph of $f(x) = 3^{x}$ , which has as its horizontal asymptote, to the right two units and down three units. Because of the latter translation, the horizontal asymptote is .

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Question

Give the vertical asymptote of the graph of the function

$g(x) = 16 \cdot 4^{x}- 3$

Answer

Since 4 can be raised to the power of any real number, the domain of is the set of all real numbers. Therefore, there is no vertical asymptote of the graph of .

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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

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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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

Find the range for,

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

Answer

$The\ graph\ of\ the\ function\ y=3^{x}-2\ has\ an\ asymptote\ at\ y=-2.$

$An\ asymptote\ on\ a\ graph\ is\ a\ value\ that\ a \ function\ approaches\ but\ never\ touches.$

$Since\ this\ graph\ approaches\ but\ never\ touches\ -2\ we\ use\ the >\ symbol.$

$\Also\ because\ the\ 3\ is\ positive\ that\ means\ the\ graph\ is\ concave\ up\ so\ >\ is\ chosen\ instead\ of\ <.$

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Question

An important part of graphing an exponential function is to find its -intercept and concavity.

Find the -intercept for

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

and determine if the graph is concave up or concave down.

Answer

$\The\ concavity\ of\ an\ exponential\ graph\ can\ be\ determined\ by\ its\ a-value.$

$This\ function\ is\ in\ the\ form\ y=a(b)^{x}+k.$

$\ So\ the\ a\ value\ is\ 2, based\ on\ the\ function\ we\ were\ given\ y=2(3)^{x}-5$

$If\ the\ a\ value\ is\ positive,\ the\ graph\ will\ be\ concave\ up$ .

$If\ the\ a\ value\ is\ negative,\ the\ graph\ will\ be\ concave\ down.$

$To\ find\ the\ y-intercept,\ plug\ in\ x=0, and\ keep\ in\ mind\ order\ of\ operations,$

$\ Exponent first, then\ multiply.$

$y=2(3)^{0}-5$

$The\ y-intercept\ is\ (0,-3).$

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Question

Give the equation of the vertical asymptote of the graph of the equation $f(x) = 2e^{x-3}+ 7$ .

Answer

Define $g(x) = e^{x}$ . In terms of , can be restated as

The graph of is a transformation of that of . As an exponential function, has a graph that has no vertical asymptote, as is defined for all real values of ; it follows that being a transformation of this function, also has a graph with no vertical asymptote as well.

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Question

Give the equation of the horizontal asymptote of the graph of the equation $f(x) = 2e^{x-3}- 7$ .

Answer

Define $g(x) = e ^{x}$ . In terms of , can be restated as

$f(x) = 2e^{x-3}- 7$ .

The graph of has as its horizontal asymptote the line of the equation . The graph of is a transformation of that of - a right shift of 3 units ( ), a vertical stretch ( ), and a downward shift of 7 units ( ). The right shift and the vertical stretch do not affect the position of the horizontal asymptote, but the downward shift moves the asymptote to the line of the equation . This is the correct response.

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