<s>[INST] Difference quotient of $f(x)= 2-6x+4x^2$ I need to find $f(a), f(a + h)$, and the difference quotient 
$$\frac {f(a + h) − f(a)}{h},$$
where $h\neq 0$ and $f(x) = 2-6x+4x^2$.
My work: 
$$f(a) = 2-6a+4a^2,\ \ f(a+h) = 2-6(a+h)+4(a+h)^2.$$
I need help on the last one: 
$$\frac {f(a+h)-f(a)}{h}.$$
My work:
\begin{split} \frac {f(a+h)-f(a)}{h} &= \frac {(a+h)-(a)}{h}\\
&= \frac {2-6a-6h+4a^2+8ah+8h^2-2+6a-4a^2}{h}\\
&=\frac {-6h+8ah+8h^2}{h}\\
&= {-6+8a+8h}
\end{split}
UPDATE: It looks like I accidentally squared the numerator. So instead it would be $\frac{6+8a+8h}{h}$. Can anyone confirm?
 [/INST] You need $\frac{f(a+h)-f(a)}{h}$, but you seem to have computed something else not involving $f$. Try this, 
\begin{equation}
\frac{f(a+h)-f(a)}{h}=\frac{(2-6(a+h)+4(a+h)^2)-(2-6a+4a^2)}{h} \\
=\frac{2-6a-6h+4a^2+8ah+4h^2-2+6a-4a^2}{h} \\
=\frac{-6h+8ah+4h^2}{h} \\
=-6+8a+4h
\end{equation}
</s>