2. [tex]f(x)=\dfrac18\left(\dfrac14\right)^{x-2}[/tex]
Domain: [tex](-\infty,\infty)[/tex], because any value of [tex]x[/tex] is allowed and gives a number [tex]f(x)[/tex].
Range: [tex](0,\infty)[/tex], because [tex]a^x>0[/tex] for any positive real [tex]a\neq0[/tex].
y-intercept: This is a point of the form [tex](0, f(0))[/tex]. So plug in [tex]x=0[/tex]; we get [tex]f(0)=\frac18\left(\frac14\right)^{-2}=\frac{4^2}8=2[/tex]. 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 [tex]y=0[/tex].
Increasing interval: Going from left to right, there is no interval on which [tex]f(x)[/tex] is increasing, since 1/4 is between 0 and 1.
Decreasing interval: Same as the domain; [tex]f(x)[/tex] is decreasing over the entire real line.
End behavior: The range tells you [tex]f(x)>0[/tex], and you know [tex]f(x)[/tex] is decreasing over its entire domain. This means that [tex]f(x)\to+\infty[/tex] as [tex]x\to-\infty[/tex], and [tex]f(x)\to0[/tex] and [tex]x\to+\infty[/tex].
3. [tex]f(x)=\left(\dfrac32\right)^{-x}-7[/tex]
Domain: Same as (2), [tex](-\infty, \infty)[/tex].
Range: We can rewrite [tex]f(x)=\left(\frac23\right)^x-7[/tex]. [tex]\left(\frac23\right)^x>0[/tex] for all [tex]x[/tex], so [tex]\left(\frac23\right)^x-7>-7[/tex] for all [tex]x[/tex]. Then the range is [tex](-7,\infty)[/tex].
y-intercept: We have [tex]f(0)=\left(\frac32\right)^0-7=1-7=-6[/tex], so the intercept is (0, -6) (or just -6).
Asymptote: [tex]y=-7[/tex]
Increasing interval: Not increasing anywhere
Decreasing interval: [tex](-\infty,\infty)[/tex]
End behavior: Similar to (2), but this time [tex]f(x)\to+\infty[/tex] as [tex]x\to-\infty[/tex] and [tex]f(x)\to-7[/tex] as [tex]x\to+\infty[/tex].
