Answer
$$
f(x)=2x^{2}-4x-5
$$
(a)
$$
\begin{split}
f(x+h) & =2 x^{2}+4 h x+2 h^{2}-4 x-4 h-5
\end{split}
$$
(b)
$$
\begin{split}
f(x+h)-f(x) = 4 h x+2 h^{2}-4h
\end{split}
$$
(c)
$$
\begin{split}
\frac{f(x+h)-f(x)}{h} = 4 x+2 h-4
\end{split}
$$
Work Step by Step
$$
f(x)=2x^{2}-4x-5
$$
(a)
$$
\begin{split}
f(x+h) & =2(x+h)^{2}-4(x+h)-5 \\
&=2\left(x^{2}+2 h x+h^{2}\right)-4 x-4 h-5 \\
&=2 x^{2}+4 h x+2 h^{2}-4 x-4 h-5
\end{split}
$$
(b)
$$
\begin{split}
f(x+h)-f(x) & = 2 x^{2}+4 h x+2 h^{2}-4 x-4 h-5 -\left(2 x^{2}-4 x-5\right) \\
& = 2 x^{2}+4 h x+2 h^{2}-4 x-4 h-5 -2 x^{2}+4 x+5 \\
& = 4 h x+2 h^{2}-4 h
\end{split}
$$
(c)
$$
\begin{split}
\frac{f(x+h)-f(x)}{h} & = \frac{4 h x+2 h^{2}-4 h}{h} \\
& = \frac{h(4 x+2 h-4)}{h} \\
& = 4 x+2 h-4
\end{split}
$$
