Question

Evaluate the function at the specified value of the independent variable and simplify.

Answer 1

$q(t)=\dfrac{5t^2+6}{t^2}\\\\1.\\q(2)=\dfrac{5\cdot2^2+6}{2^2}=\dfrac{5\cdot4+6}{4}=\dfrac{20+6}{4}=\dfrac{26}{4}=7.5\\\\2.\\q(0)=\dfrac{5\cdot0^2+6}{0^2}=\dfrac{6}{0}-unde fined\\\\3.\\q(-x)=\dfrac{5\cdot(-x)^2+6}{(-x)^2}=\dfrac{5x^2+6}{x^2}$