Question

Can y’all help me answer these questions

Answer 1

2. $f(x)=\dfrac18\left(\dfrac14\right)^{x-2}$

Domain: $(-\infty,\infty)$, because any value of $x$ is allowed and gives a number $f(x)$.

Range: $(0,\infty)$, because $a^x>0$ for any positive real $a\neq0$.

y-intercept: This is a point of the form $(0, f(0))$. So plug in $x=0$; we get $f(0)=\frac18\left(\frac14\right)^{-2}=\frac{4^2}8=2$. So the intercept is (0, 2), or just 2. (Interestingly, you didn't get marked wrong for that...)

Asymptote: This can be deduced from the range; the asymptote is the line $y=0$.

Increasing interval: Going from left to right, there is no interval on which $f(x)$ is increasing, since 1/4 is between 0 and 1.

Decreasing interval: Same as the domain; $f(x)$ is decreasing over the entire real line.

End behavior: The range tells you $f(x)>0$, and you know $f(x)$ is decreasing over its entire domain. This means that $f(x)\to+\infty$ as $x\to-\infty$, and $f(x)\to0$ and $x\to+\infty$.

3. $f(x)=\left(\dfrac32\right)^{-x}-7$

Domain: Same as (2), $(-\infty, \infty)$.

Range: We can rewrite $f(x)=\left(\frac23\right)^x-7$. $\left(\frac23\right)^x>0$ for all $x$, so $\left(\frac23\right)^x-7>-7$ for all $x$. Then the range is $(-7,\infty)$.

y-intercept: We have $f(0)=\left(\frac32\right)^0-7=1-7=-6$, so the intercept is (0, -6) (or just -6).

Asymptote: $y=-7$

Increasing interval: Not increasing anywhere

Decreasing interval: $(-\infty,\infty)$

End behavior: Similar to (2), but this time $f(x)\to+\infty$ as $x\to-\infty$ and $f(x)\to-7$ as $x\to+\infty$.