Answer
$\dfrac{f(x+h)-f(x)}{h}=2x+h-2$
$\dfrac{f(x)-f(a)}{x-a}=x+a-2$
Work Step by Step
$f(x)=x^{2}-2x$
$\textbf{Evaluate}$ $\dfrac{f(x+h)-f(x)}{h}$
First, substitute $x$ by $x+h$ in the given function and simplify to find $f(x+h)$:
$f(x+h)=(x+h)^{2}-2(x+h)=x^{2}+2xh+h^{2}-2x-2h$
Substitute $f(x+h)$ and $f(x)$ into the difference quotient formula and simplify:
$\dfrac{f(x+h)-f(x)}{h}=\dfrac{x^{2}+2xh+h^{2}-2x-2h-(x^{2}-2x)}{h}=...$
$...=\dfrac{x^{2}+2xh+h^{2}-2x-2h-x^{2}+2x}{h}=...$
$...=\dfrac{2xh+h^{2}-2h}{h}=...$
Take out common factor $h$ from the numerator and simplify again:
$...=\dfrac{h(2x+h-2)}{h}=2x+h-2$
$\textbf{Evaluate}$ $\dfrac{f(x)-f(a)}{x-a}$
Substitute $x$ by $a$ in $f(x)$ to find $f(a)$:
$f(a)=a^{2}-2a$
Substitute $f(a)$ and $f(x)$ into the difference quotient formula and simplify:
$\dfrac{f(x)-f(a)}{x-a}=\dfrac{x^{2}-2x-(a^{2}-2a)}{x-a}=\dfrac{x^{2}-2x-a^{2}+2a}{x-a}=...$
Factor the numerator by grouping. The first group is a difference of squares and the second group can be factored taking out common factor $2$:
$...=\dfrac{(x^{2}-a^{2})-(2x-2a)}{x-a}=\dfrac{(x-a)(x+a)-2(x-a)}{x-a}=...$
Take out common factor $x-a$ from the numerator and simplify again:
$...=\dfrac{(x-a)(x+a-2)}{x-a}=x+a-2$
