If $f(x+1) = x^2 + 3x +5$, then find $f(x)$ A challenge problem from precalc class. I don't know what to do with the one from $f(x+1)$, it doesn't factor well, I could do completing the square but then how do I find $f(x)$? Just stuck on this. 

Just to be different. 
$$f(x+1) = x^2 + 3x + 5$$
\begin{align}
   f(x) - f(x+1)
   &= f((x-1)+1) - f(x) \\
   &= (x-1)^2 + 3(x-1) + 5) - (x^2 + 3x + 5) \\
   &= ((x-1)^2 - x^2) + (3(x-1) - 3x) + (5-5) \\
   &= (x-1-x)(x-1+x) - 3\\
   &= (-1)(2x-1) - 3 \\
   &= -2x - 2
\end{align}
So $f(x) = f(x+1) - 2x - 2 = x^2 + x + 3$
proof by induction
It is not unreasonable to suppose that $f(x) = x^2 + ax + b$ for some real numbers $a$ and $b$. We can use inductive reasoning to prove that this is the case and to find the values of $a$ and $b$ at the same time.
Our hypothesis will be $f(x) = x^2 + ax + b$ for some real numbers $a$ and $b$. 
$f(0) = f(-1+1) = (-1)^2 + 3(-1) + 5 = 3$
By our hypothesis, $f(0) = b$. Hence $b = 3$
So our hypothesis is now $f(x) = x^2 + ax + 3$
Again, by our hypothesis, $f(x+1) = (x+1)^2 + a(x+1) + 3$
Hence
\begin{align}
   x^2 + 3x + 5 &= (x+1)^2 + a(x+1) + 3 \\
   x^2 + 3x + 5 &= (x^2+2x+1) + (ax + a) + 3 \\
   x^2 + 3x + 5 &= x^2 + (2 + a)x + (a + 4) \\
\end{align}
And we see that this is true when $a = 1$.
Hence, by mathematical induction,  $f(x) = x^2 + x + 3$.
Now that I look at this, I realize that I have only shown that $f(n) = n^2 + n + 3$ for $n = 0, 1, 2, \dots$. It is true however that three points uniquely determine a parabola. And we have agreement at an infinite number of points. So this is still a proof.
