## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$2x+h-1$
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 $2x+h-1$.