algebra precalculus - $f$ is a function so that $f(ab + 1) = af(b) − f(a) + 6$ for all real numbers $a$ and $b$

If a function $f$ exists so that $f(ab + 1) = af(b) − f(a) + 6$ for all real numbers $a$ and $b$. What are all possible functions $f$ that satisfy this equation and also prove that there are no other functional solutions that make the equation true? Any help is appreciated

Answer

Put $b =0$ to get $f(1)=af(0)-f(a)+6$. Solve this for $f(a)$.

You get $f(x)=cx+d$ for some $c$ and $d$. Now go back to the original equation to see what you can say about $c$ and $d$.

The answer is $f(x)=2x+2$.