Answer
(a) $q(2)=\frac{11}{4}$
(b) It is not possible.
(c) $q(-x)=\frac{2x^2+3}{x^2}$
Work Step by Step
$q(t)=\frac{2t^2+3}{t^2}$
(a) $q(2)=\frac{2(2)^2+3}{2^2}=\frac{2(4)+3}{4}=\frac{11}{4}$
(b) $q(0)=\frac{2(0)^2+3}{0^2}=\frac{3}{0}$ (It is not possible to divide by $0$)
(c) $q(-x)=\frac{2(-x)^2+3}{(-x)^2}=\frac{2x^2+3}{x^2}$
