How to graph an exponential function - Geometry

Card 0 of 104

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

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

Tap to see back →

Set and solve for :

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

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

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

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

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.

Tap to see back →

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:

$x \approx \ln 5.333 \approx 1.7$

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.

Tap to see back →

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}$

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

$y \approx 14.3$

Give the -intercept of the graph of the function

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

Round to the nearest tenth, if applicable.

Tap to see back →

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$

$\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 .

Give the -intercept of the graph of the function

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

Round to the nearest hundredth, if applicable.

Tap to see back →

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.

Give the horizontal asymptote of the graph of the function

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

Tap to see back →

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 .

Give the vertical asymptote of the graph of the function

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

Tap to see back →

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 .

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to see back →

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 )$ .

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to see back →

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.

Define a function as follows:

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

Give the horizontal aysmptote of the graph of .

Tap to see back →

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.

Define a function as follows:

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

Give the vertical aysmptote of the graph of .

Tap to see back →

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.

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to see back →

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 .

Define a function as follows:

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

Give the -intercept of the graph of .

Tap to see back →

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 .

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 see back →

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.

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 see back →

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.

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

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

Evaluate .

Tap to see back →

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

Find the range for,

Tap to see back →

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

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

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

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

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

Find the -intercept for

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

Tap to see back →

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

$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$

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

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

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

$\ Exponent first, then\ multiply.$

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

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

Tap to see back →

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.

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

Tap to see back →

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.