# Converting quadratic equation from normal form to vertex form

The online calculator calculates the solutions of quadratic equations with solution steps.

Parabola Plotter

## 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:

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

## Zeros from vertex form

From the vertex form of the quadratic function it is easy to calculate the zeros of the function.

Starting from the vertex form

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

the condition for zeros is that the function is zero

$0={\left(x-{x}_{S}\right)}^{2}+{y}_{S}$

and reshaping yields

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

$x-{x}_{S}=±\sqrt{-{y}_{S}}$

and finally to the zeros

${x}_{1,2}={x}_{S}±\sqrt{-{y}_{S}}$

## 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 of the parabola 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 expansion and application of the first binomial:

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

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

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

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

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

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

 p = q =

## 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:

 a = b = c =