Solve the difference quotient 
So I tried a variety of answers and they all came up incorrect. I guess I don't understand difference quotient because I feel like it's pointless and a slow way to do the problem. That being said... can't get the answer
 A: You are expected to write
$$\frac{f(x)-f(4)}{x-4}=\frac{\frac{x+6}{x+4}-\frac{10}{8}}{x-4},\tag{1}$$
and then simplify.
We first simplify the numerator. Bringing it to the common denominator $(x+4)(8)$, we get
$$  \frac{x+6}{x+4}-\frac{10}{8}=\frac{(x+6)(8)-(x+4)(10)}{(x+4)(8)}=\frac{-2x+8}{(x+4)(8)}.$$
This simplifies to $\frac{-x+4}{(x+4)(4)}$.
Dividing by $x-4$, we find that if $x\ne 4$, then
$$\frac{f(x)-f(4)}{x-4}=-\frac{1}{(x+4)(4)}.$$
Remarks: The algebra is somewhat simpler if first we rewrite $\frac{x+6}{x+4}$ as $1+\frac{2}{x+4}$. 
From the answer you got, it looks as if you evaluated the difference quotient correctly at $x=4$. That is not what the question asked for: you did extras work!
A: Evaluate $f(4)$:
$$f(4) = \frac{4+6}{4+4}=\frac{10}{8}$$
Plug in what you know into the difference quotient:
$$\frac{\frac{x+6}{x+4}-\frac{10}{8}}{x-4}$$
Simplifying the numerator:
$$\frac{x+6}{x+4}-\frac{10}{8}=\frac{8(x+6)-10(x+4)}{8(x+4)}=\frac{8-2x}{8(x+4)}=\frac{-2(x-4)}{8(x+4)}=-\frac{(x-4)}{4(x+4)}$$
$$\frac{-\frac{x-4}{4(x+4)}}{x-4}=-\frac{x-4}{4(x+4)(x-4)}=-\frac{1}{4(x+4)}$$
