Putting a quadratic into standard form

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Let's start with the function

\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).

Example from Stewart

Mathematica:

  1. Mathematica File confirming the formula for the y-value of the vertex.
  2. Animation of Standard Form.
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