# Putting a quadratic into standard form

${\displaystyle \left.f(x)=ax^{2}+bx+c\right.}$

where ${\displaystyle a\neq {0}}$ (otherwise we'd have a linear function).

Now we factor out an ${\displaystyle a}$:

${\displaystyle \left.f(x)=a\left(x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}\right)\right.}$

We want to create a perfect square, by replacing that term

${\displaystyle \left.x^{2}+{\frac {b}{a}}x\right.}$

How do we do that?

Well, thinking backwards is often useful. If we consider a perfect square and expand it,

${\displaystyle \left.(x+d)^{2}=x^{2}+2dx+d^{2}\right.}$,

we see that we can rewrite that as

${\displaystyle \left.(x+d)^{2}-d^{2}=x^{2}+2dx\right.}$.

We've solved for the quadratic piece and the linear piece. Now we set this equal to the expression we want to replace.

Setting

${\displaystyle \left.2d={\frac {b}{a}}\right.}$

or

${\displaystyle d=\left.{\frac {b}{2a}}\right.}$

we can write

${\displaystyle \left.(x+\left.{\frac {b}{2a}}\right.)^{2}-\left(\left.{\frac {b}{2a}}\right.\right)^{2}=x^{2}+2\left(\left.{\frac {b}{2a}}\right.\right)x=x^{2}+\left.{\frac {b}{a}}\right.x\right.}$.

from which we arrive at

${\displaystyle \left.f(x)=a\left((x+{\frac {b}{2a}})^{2}-({\frac {b}{2a}})^{2}+{\frac {c}{a}}\right)\right.}$

Finally, following a little simplification, we can write ${\displaystyle f(x)}$ in standard form as

${\displaystyle \left.f(x)=a(x-{\frac {-b}{2a}})^{2}+{\frac {4ac-b^{2}}{4a}}\right.}$

or

${\displaystyle \left.f(x)=a(x-{\frac {-b}{2a}})^{2}+c-{\frac {b^{2}}{4a}}\right.}$

The former has the advantage of featuring the discriminant, from the quadratic formula. The simplest way of finding the constant is by evaluating

${\displaystyle \left.f\left({\frac {-b}{2a}}\right)\right.}$

From this form, we can see that the maximum or minimum of the function occurs at

${\displaystyle \left.x=-{\frac {b}{2a}}\right.}$

because this is the value at which the squared term is zero. The value of the function at ${\displaystyle x=-{\frac {b}{2a}}}$ is

${\displaystyle \left.{\frac {4ac-b^{2}}{4a}}\right.}$

These two special values together are called the vertex: ${\displaystyle \left(-{\frac {b}{2a}},{\frac {4ac-b^{2}}{4a}}\right)}$.

Mathematica: