$$\frac{f(x+h)-f(x)}{h}=\frac{\left[(x+h)^2-3(x+h)+1\right]-\left[x^2-3x+1\right]}{h}\\[0.45 in] = \frac{\left[x^2+2xh+h^2-3x-3h+1\right]-\left[x^2-3x+1\right]}{h}\\[0.45 in] = \frac{x^2+2xh+h^2-3x-3h+1-x^2+3x-1}{h}\\[0.45 in] = \frac{2xh+h^2-3h}{h} = \frac{h(2x+h-3)}{h}\\[0.45 in] =2x+h-3$$