Answer
$$f'\left( x \right) = - 8x + 9,\,\,\,\,f'\left( { - 2} \right) = 25,\,\,\,\,f'\left( 0 \right) = 9,{\text{ and }}f'\left( 3 \right) = - 15$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = - 4{x^2} + 9x + 2 \cr
& {\text{Using the definition of the derivative}}: \cr
& {\text{Find }}f\left( {x + h} \right) \cr
& f\left( {x + h} \right) = - 4{\left( {x + h} \right)^2} + 9\left( {x + h} \right) + 2 \cr
& f\left( {x + h} \right) = - 4\left( {{x^2} + 2xh + {h^2}} \right) + 9\left( {x + h} \right) + 2 \cr
& f\left( {x + h} \right) = - 4{x^2} - 8xh - 4{h^2} + 9x + 9h + 2 \cr
& {\text{Find and simplify }}f\left( {x + h} \right) - f\left( x \right) \cr
& f\left( {x + h} \right) - f\left( x \right) = - 4{x^2} - 8xh - 4{h^2} + 9x + 9h + 2 - \left( { - 4{x^2} + 9x + 2} \right) \cr
& f\left( {x + h} \right) - f\left( x \right) = - 4{x^2} - 8xh - 4{h^2} + 9x + 9h + 2 + 4{x^2} - 9x - 2 \cr
& f\left( {x + h} \right) - f\left( x \right) = - 8xh - 4{h^2} + 9h \cr
& {\text{Divide by }}h{\text{ to get }}\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{ - 8xh - 4{h^2} + 9h}}{h} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = - 8x - 4h + 9 \cr
& {\text{Let }}h \to 0 \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \left( { - 8x - 4h + 9} \right) \cr
& {\text{Then}}{\text{,}} \cr
& f'\left( x \right) = - 8x - 4\left( 0 \right) + 9 \cr
& f'\left( x \right) = - 8x + 9 \cr
& \cr
& {\text{Find }}f'\left( { - 2} \right),f'\left( 0 \right){\text{ and }}f'\left( 3 \right) \cr
& f'\left( { - 2} \right) = - 8\left( { - 2} \right) + 9 = 25 \cr
& f'\left( 0 \right) = - 8\left( 0 \right) + 9 = 9 \cr
& f'\left( 3 \right) = - 8\left( 3 \right) + 9 = - 15 \cr} $$
