How do you express the quadratic equation in terms of mu and variance? The Wikipedia article for the Normal distribution says that it's more common to express the quadratic function $f(x) = ax^2+bx+c$ in terms of $\mu$ and $\sigma^2$ what is the transformation that I need to do to eventually solve $\mu = -\frac{b}{2a}$ and $\sigma^2 = -\frac{1}{2a}$
I know I can equate $ax^2 + bx +c =\frac{-(x-\mu)^2}{2\sigma^2}$, but I don't see how the right term is derived.
 A: Complete the square:
$$
  ax^2+bx+c=a\left(x^2+\frac{b}{a}x\right)+c=a\left(x^2+2\frac{b}{2a}x +\frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c
$$
$$
= a\left(x+\frac{b}{2a}\right)^2 -\frac{b^2}{4a} + c \, .
$$
Hence,
$$
  e^{ax^2+bx+c} = A\,e^{a\left(x+\frac{b}{2a}\right)^2} \, ,
$$
in which $A=e^{- \frac{b^2}{4a} + c}$, and this is a normal density of the form
$$
  \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2}\left(x-\mu \right)^2}
$$
if and only if $\mu = -\frac{b}{2a}$, $-\frac{1}{2\sigma^2}=a$, and $\frac{1}{\sqrt{2\pi}\sigma} = A$.
You end with just two parameters, $\mu$ and $\sigma$, instead of three, because you need a constraint that guarantees that $e^{ax^2+bx+c}$ integrates to one (it is a density). In other words, you already know that $c$ is determined from $a$ and $b$ by $c=-\log\left(\int_{-\infty}^\infty e^{ax^2+bx}\, dx\right)$.
A: You algebraically expand $-(x-\mu)^2/(2\sigma^2)$ and equate the coefficients.  Notice $a<0$.  This works in yhr opposite direction as in Zen's solution.  But as he points out, there are three equations in two unknowns.  But the constraint that the density integrates to 1 does reduce it to 2 equations in two unknowns.  His way fills in the details.
