Answer
$2x+h-1$
Work Step by Step
Given $f(x)=x^2-x+4$
The difference quotient of a function $f(x)$ can be obtained by $\frac{f(x+h)-f(x)}{h}$
provided $h\ne0$, hence:
$\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^2-(x+h)+4-(x^2-x+4)}{h}$
$=\frac{x^2+2xh+h^2-(x+h)+4-(x^2-x+4)}{h}$
$=\frac{x^2+2xh+h^2-x-h+4-x^2+x-4}{h}$
$=\frac{2xh+h^2-h}{h}$
$=\frac{h(2x+h-1)}{h}$
$=2x+h-1$
Thus, the difference quotient is $2x+h-1$.
