Answer
(a) $\frac{x^2+6x+8}{(x+3)^2}$
(b) $-2,-4$
(c) vertical asymptote $x=-3$, horizontal asymptote $y=1$
Work Step by Step
(a) translating the graph of $y=-\frac{1}{x^2}$ to the left 3 units and up 1 unit will result an equation of $f(x)=-\frac{1}{(x+3)^2}+1=\frac{x^2+6x+8}{(x+3)^2}$
(b) The zero(s) of $f(x)$ can be found by letting $f(x)=0$ or $x^2+6x+8=0$ which gives $x=-2,-4$
(c) We can identify a vertical asymptote $x=-3$ and a horizontal asymptote $y=1$
