Answer
$$
f(x)=-4 x^{2}+11 x
$$
Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0}\frac{-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x) }{h}\\
&=\lim\limits_{h \to 0}\frac{-8 x h-4 h^{2}+11 h}{h}\\
&=\lim\limits_{h \to 0}(-8 x -4 h+11)\\
&=-8 x +11.
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=-8 (2) +11=-5,
$$
$$
f^{\prime}(16)=-8 (16) +11=-117
$$
and
$$
f^{\prime}(-3)=-8 (-3) +11=35
$$
Work Step by Step
$$
f(x)=-4 x^{2}+11 x
$$
Go through the four steps to find $f^{\prime}(x)$
Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$
$$
f(x +h)=-4 (x +h)^{2}+11 (x +h)
$$
Step 2
$$
f(x +h) f(x)=-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x)
$$
Step 3
$$
\frac{f(x +h)- f(x)}{h}= \frac{-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x) }{h}
$$
Step 4 Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0}\frac{-4 (x +h)^{2}+11 (x +h)-(-4 x^{2}+11 x) }{h}\\
&=\lim\limits_{h \to 0}\frac{-8 x h-4 h^{2}+11 h}{h}\\
&=\lim\limits_{h \to 0}(-8 x -4 h+11)\\
&=-8 x +11.
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=-8 (2) +11=-5,
$$
$$
f^{\prime}(16)=-8 (16) +11=-117
$$
and
$$
f^{\prime}(-3)=-8 (-3) +11=35.
$$
