## Vertex form

The vertex form of the square function is:

$y={\left(x-{x}_{S}\right)}^{2}+{y}_{S}$

Where x _{ S } and y _{ S } are the x and y coordinates of the vertex of the parabola. The vertex is the minimum or maximum of the function, depending on whether the parabola is up or down.

## Basic form

In the basic form, the coefficient before x ^{ 2 is 1.}

Basic form of the quadratic function with the constant coefficients p and q:

$y={x}^{2}+px+q$

If the square function is in basic form, the vertex is given by:

${x}_{S}=-\frac{p}{2}$

${y}_{S}=-{\left(\frac{p}{2}\right)}^{2}+q$

Transformation from the basic form to the vertex form with quadratic addition and application of the first binomial:

${x}^{2}+px+q=$

${x}^{2}+px+{\left(\frac{p}{2}\right)}^{2}-{\left(\frac{p}{2}\right)}^{2}+q=$

${\left(x+\frac{p}{2}\right)}^{2}-{\frac{p}{2}}^{2}+q=$

${\left(x-\frac{-p}{2}\right)}^{2}-{\frac{p}{2}}^{2}+q$

## Calculator for the conversion from the basic form to the vertex form

### Input the coefficients p and q of the quadratic equation:

## General form

General form of the quadratic function with the constant coefficients a, b, and c:

$y=a{x}^{2}+bx+c$

If the quadratic function is in the general form the vertex is given by:

${x}_{S}=-\frac{b}{2a}$

${y}_{S}=-\frac{{b}^{2}}{4a}+c$

Transformation from the general form to the vertex form with quadratic complement and application of the first binomial:

$a{x}^{2}+bx+c=$

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

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

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

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

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

## Calculator for the conversion from the general form to the vertex form

### Input the coefficients a, b and c of the quadratic function:

## Vertex of the parabola

The determination of the vertex of a quadratic function is performed by deriving the function. The condition for an extremum is that the first derivative of the function vanishes. For a square function this is sufficient for a minimum or maximum.

The derivation of the general form is:

$y\prime =2ax+b$

The condition for the vertex is that the derivative is 0.

$2ax+b=0$

Resolving yields the x coordinate of the vertex:

${x}_{S}=-\frac{b}{2a}$

Inserting into the general quadratic function yields the y-coordinate of the vertex:

${y}_{S}=-\frac{{b}^{2}}{4a}+c$