 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 a parbola in basic form Vertex of a parbola in general form 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 =