The graph of $y=g(-x)$ passes through the point $(2,1)$ . If $g$ is an exponential function, which of the following could define $g$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $g(x)=\left(\frac{1}{2}\right)^{x}-3$ $\newline$ (B) $g(x)=16\left(\frac{3}{4}\right)^{x}-8$ $\newline$ (C) $g(x)=2^{x}-5$ $\newline$ (D) $g(x)=18(3)^{x}-3$

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Algebra 1

Compare linear and exponential growth

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Q. The graph of

y=g(-x)

passes through the point

(2,1)

. If

g

is an exponential function, which of the following could define

g

\newline

Choose

1

answer:

\newline

(A)

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

\newline

(B)

g(x)=16\left(\frac{3}{4}\right)^{x}-8

\newline

(C)

g(x)=2^{x}-5

\newline

(D)

g(x)=18(3)^{x}-3

Given Point Analysis: We are given that the graph of $y=g(-x)$ passes through the point $(2,1)$ . This means that when $x=2$ , $g(-x)=g(-2)=1$ . We will use this information to test each of the given options for $g(x)$ to see which one satisfies this condition.
Option (A) Evaluation: Let's start with option (A) $g(x)=\left(\frac{1}{2}\right)^x-3$ . We substitute $x$ with $-2$ to get $g(-2)=\left(\frac{1}{2}\right)^{-2}-3$ . Calculating this gives $g(-2)=4-3=1$ .
Option (B) Evaluation: Since $g(-2)=1$ for option (A), this option satisfies the condition that the graph of $y=g(-x)$ passes through the point $(2,1)$ . We should check the other options to ensure there is no other function that also satisfies the condition.
Option (C) Evaluation: Now let's check option (B) $g(x)=16\left(\frac{3}{4}\right)^x-8$ . We substitute $x$ with $-2$ to get $g(-2)=16\left(\frac{3}{4}\right)^{-2}-8$ . Calculating this gives $g(-2)=16\times\left(\frac{16}{9}\right)-8=\frac{256}{9}-8$ which is not equal to $1$ .
Option (D) Evaluation: Option (B) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (C).
Option (D) Evaluation: Option (B) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (C).For option (C) $g(x)=2^x-5$ , we substitute $x$ with $-2$ to get $g(-2)=2^{-2}-5$ . Calculating this gives $g(-2)=\frac{1}{4}-5$ which is not equal to $1$ .
Option (D) Evaluation: Option (B) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (C).For option (C) $g(x)=2^x-5$ , we substitute $x$ with $-2$ to get $g(-2)=2^{-2}-5$ . Calculating this gives $g(-2)=\frac{1}{4}-5$ which is not equal to $1$ .Option (C) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (D).
Option (D) Evaluation: Option (B) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (C).For option (C) $g(x)=2^x-5$ , we substitute $x$ with $-2$ to get $g(-2)=2^{-2}-5$ . Calculating this gives $g(-2)=\frac{1}{4}-5$ which is not equal to $1$ .Option (C) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (D).Finally, let's check option (D) $1$ $0$ . We substitute $x$ with $-2$ to get $1$ $3$ . Calculating this gives $1$ $4$ which is not equal to $1$ .
Option (D) Evaluation: Option (B) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (C).For option (C) $g(x)=2^x-5$ , we substitute $x$ with $-2$ to get $g(-2)=2^{-2}-5$ . Calculating this gives $g(-2)=\frac{1}{4}-5$ which is not equal to $1$ .Option (C) does not satisfy the condition since $g(-2)$ is not equal to $1$ . We will move on to option (D).Finally, let's check option (D) $1$ $0$ . We substitute $x$ with $-2$ to get $1$ $3$ . Calculating this gives $1$ $4$ which is not equal to $1$ .Option (D) does not satisfy the condition since $g(-2)$ is not equal to $1$ . Therefore, the only option that satisfies the condition that the graph of $1$ $8$ passes through the point $1$ $9$ is option (A).