Question

Can y’all help me answer these questions

Answer 1

2.
`f(x)=\frac18\left(\frac14\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\\eq0`.

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(\frac32\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`.