## Precalculus (6th Edition) Blitzer

$2x+h+3$
Step 1. Given $f(x)=x^2+3x+2$, we have $f(x+h)=(x+h)^2+3(x+h)+2=x^2+2xh+h^2+3x+3h+2$ Step 2. We have $f(x+h)-f(x)=2xh+h^2+3h$ Step 3. Thus $\frac{f(x+h)-f(x)}{h}=2x+h+3$