# Putting a quadratic into standard form $\left.f(x)=ax^2+bx+c\right.$

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

Now we factor out an a: $\left.f(x)=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\right.$

We want to create a perfect square, by replacing that term $\left.x^2+\frac{b}{a}x\right.$

How do we do that?

Well, thinking backwards is often useful. If we consider a perfect square and expand it, $\left.(x+d)^2=x^2+2dx+d^2\right.$,

we see that we can rewrite that as $\left.(x+d)^2-d^2=x^2+2dx\right.$.

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

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

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

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

from which we arrive at $\left.f(x)=a\left((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}\right)\right.$

Finally, following a little simplification, we can write f(x) in standard form as $\left.f(x)=a(x-\frac{-b}{2a})^2+\frac{4ac-b^2}{4a}\right.$

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

The former has the advantage of featuring the discriminant, from the quadratic formula. The simplest way of finding the constant is by evaluating $\left.f\left(\frac{-b}{2a}\right)\right.$

From this form, we can see that the maximum or minimum of the function occurs at $\left.x=-\frac{b}{2a}\right.$

because this is the value at which the squared term is zero. The value of the function at $x=-\frac{b}{2a}$ is $\left.\frac{4ac-b^2}{4a}\right.$

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

Mathematica: