Answer
(a) $q(0)=-\frac{1}{9}$
(b) It is not possible.
(c) $q(y+3)=\frac{1}{y^2+6y}$
Work Step by Step
$q(x)=\frac{1}{x^2-9}$
(a) $q(0)=\frac{1}{0^2-9}=-\frac{1}{9}$
(b) $q(3)=\frac{1}{3^2-9}=\frac{1}{0}$ (it is not possible to divide by $0$)
(c) $q(y+3)=\frac{1}{(y+3)^2-9}=\frac{1}{y^2+6y+9-9}=\frac{1}{y^2+6y}$
