How to graph an exponential function - ACT Math

Card 0 of 54

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:

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

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

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

$y \approx 14.3$

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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:

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

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

$3e^{x} = 16$

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

$e^{x} = \approc 5.3333$

$x \approx \ln 5.333 \approx 1.7$

Compare your answer with the correct one above

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:

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

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

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

$y \approx 14.3$

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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:

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

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

$3e^{x} = 16$

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

$e^{x} = \approc 5.3333$

$x \approx \ln 5.333 \approx 1.7$

Compare your answer with the correct one above

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:

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

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

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

$y \approx 14.3$

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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:

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

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

$3e^{x} = 16$

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

$e^{x} = \approc 5.3333$

$x \approx \ln 5.333 \approx 1.7$

Compare your answer with the correct one above

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:

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

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

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

$y \approx 14.3$

Compare your answer with the correct one above

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