#### Answer

$ f'(3)=-11$
Equation of the tangent line:
$ y=-11t+18$

#### Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $ a=3$. Therefore, $ f'(3)=\lim\limits_{h \to 0}\frac{f(3+h)-f(3)}{h}=\lim\limits_{h \to 0}\frac{[(3+h)-2(3+h)^{2}]-[3-2(3)^{2}]}{h}$
$=\lim\limits_{h \to 0}\frac{(3+h-2\times3^{2}-2h^{2}-2\times6h)-(-15)}{h}=\lim\limits_{h \to 0}\frac{-2h^{2}-11h}{h}=\lim\limits_{h \to 0}(-2h-11)=-11$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(t-a)$
Knowing that $ f(a)=f(3)=-15,f'(a)=f'(3)=-11$ and $ a=3$, we obtain the equation of the tangent line through $ a $ as below.
$ y-(-15)=-11(t-3)$
Or $ y=-11t+18$