If $f(x)=x^2-3x+2$, then $f(x+1)-f(x)$ equals:

Substituting $x+1$ gives $f(x+1)=(x+1{)}^2-3(x+1)+2=x^2+2x+1-3x-3+2=x^2-x$. Subtracting, $f(x+1)-f(x)=(x^2-x)-(x^2-3x+2)=2x-2$. (Forgetting to subtract the constant 2 leaves $2x$, and −2 is only the value at $x=0$.) Check at $x=3:f(4)-f(3)=6-2=4=2(3)-2$.