Answer
(a) See graph.
(b) $(-\infty,3)U(3,\infty)$, $(1,\infty)$
(c) $x=3$, $y=1$
Work Step by Step
(a) To obtain the graph of $G(x)=1+\frac{2}{(x-3)^2}$ from $y=\frac{1}{x^2}$, shift the curve 3 units to the right, stretch vertically by a factor of 2, then shift 1 unit up. See graph.
(b) Based on the graph, we can find the domain as $(-\infty,3)U(3,\infty)$ and range as $(1,\infty)$
(c) We can identify the vertical asymptote $x=3$, horizontal asymptote $y=1$, or oblique asymptote $none$.
