A particular function is decreasing and has the asymptote $$ y=2 $$. Which of the following could be the function? a. $$ f(x)=\frac{1}{2}(5)^{x}+2 $$ b. $$ f(x)=-2\left(\frac{1}{5}\right)^{x}+2 $$ f. $$ f(x)=-\frac{1}{4}\left(\frac{1}{5}\right)^{x}+2 $$ d. $$ f(x)=-3(5)^{x}+2 $$

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1. We are given a function in the form $$ f(x)=a\cdot b^x+2 $$ and we know that the horizontal asymptote is $$ y=2 $$. This tells us the constant term is 2. 2. For the function to be decreasing, its derivative $$ f'(x) $$ must be negative for all $$ x $$. Since $$ f'(x)=a\cdot b^x\ln (b), $$ and $$ b^x>0 $$ for all $$ x $$, the sign of $$ f'(x) $$ depends on the product $$ a\ln(b) $$. We require: $$ a\ln(b) < 0. $$ 3. Now, check each option: **Option a.** $$ f(x)=\frac{1}{2}(5)^x+2 $$ Here, $$ a=\frac{1}{2}>0 $$ and $$ b=5>1 $$ so $$ \ln(5)>0 $$. Thus, $$ a\ln(b)=\frac{1}{2}\ln(5)>0, $$ meaning the function is increasing. **Option b.** $$ f(x)=-2\left(\frac{1}{5}\right)^x+2 $$ Here, $$ a=-2<0 $$ and $$ b=\frac{1}{5}<1 $$ so $$ \ln\left(\frac{1}{5}\right)<0 $$. The product is $$ a\ln(b)=-2\cdot\ln\left(\frac{1}{5}\right)=-2\cdot(-\ln 5)=2\ln 5>0, $$ so the function is increasing. **Option f.** $$ f(x)=-\frac{1}{4}\left(\frac{1}{5}\right)^x+2 $$ Similarly, $$ a=-\frac{1}{4}<0 $$ and $$ b=\frac{1}{5}<1 $$ so $$ \ln\left(\frac{1}{5}\right)<0 $$. Thus, $$ a\ln(b)=-\frac{1}{4}\cdot\ln\left(\frac{1}{5}\right)=-\frac{1}{4}\cdot(-\ln 5)=\frac{\ln5}{4}>0, $$ and the function is increasing. **Option d.** $$ f(x)=-3(5)^x+2 $$ Here, $$ a=-3<0 $$ and $$ b=5>1 $$ so $$ \ln(5)>0 $$. Therefore, $$ a\ln(b)=-3\ln(5)<0, $$ ensuring that $$ f(x) $$ is decreasing. 4. The only function both decreasing and with horizontal asymptote $$ y=2 $$ is option d. Option d, $$ f(x)=-3(5)^x+2 $$, is the function that is decreasing and has the horizontal asymptote $$ y=2 $$.

A particular function is decreasing and has the
asymptote . Which of the following could be
the function?
a.
b.
f.
d.

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