The calculator determines the vertex shape of the quadratic function step by step.

The general quadratic function

$y\left(x\right)=a{x}^{2}+bx+c$

is converted into the vertex form

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

Enter the coefficients a, b and c of the quadratic function:

Conversion to vertex form with quadratic addition:

The result is the vertex shape:

The vertex form of the square function is:

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

or if the square funktion is in basic form with a=1:

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

Where x_{V} and y_{V} 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.

Vertex of a quadratic function in p,q-Form

Vertex of a quadratic function in general form

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.

Start point is the general quadratic function:

$y\left(x\right)=a{x}^{2}+bx+c$

The derivative of the general form is:

$y\prime =2ax+b$

The condition for the vertex is that the derivative is 0. That means the following equation is valid:

$2ax+b=0$

Resolving yields the x coordinate of the vertex:

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

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

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

From the second derivative of the quadratic function follows wether the vertex is a maximum or minimum. The second derivative is:

$y\prime \prime =2a$

So for a > 0 the vertex is a minimum value of the parabola and for a < 0 a maximum value.

In the basic form, the coefficient before x^{2} is equal to 1. The basic form of the quadratic function with the constant coefficients p and q is

$y\left(x\right)={x}^{2}+px+q$

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

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

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

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

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

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

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

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

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

Input the coefficients p and q of the quadratic equation:

Conversion to the vertex form with quadratic expansion:

Standard 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 standard form the vertex is given by:

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

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

Transformation from the standard form to the vertex form with quadratic expansion and application of the first binomial:

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

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

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

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

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

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

$a{\left(x-{x}_{V}\right)}^{2}+{y}_{V}$

Conversion of the vertex form of the quadratic function into the standard form.

Starting point is the vertex form

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

Resolving the square results in:

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

Multiplying out the bracket results in:

$a{x}^{2}-2ax{x}_{V}+a{{x}_{V}}^{2}+{y}_{V}=$

Insertion of x_{V} and y_{V} results:

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

Shortening results in:

$a{x}^{2}+bx+\frac{{b}^{2}}{4a}-\frac{{b}^{2}}{4a}+c=$

The summands cancel each other out and the general quadratic function follows:

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

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

Starting from the vertex form

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

the condition for zeros is that the function is zero

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

and reshaping yields

${\left(x-{x}_{V}\right)}^{2}=-\frac{{y}_{V}}{a}$

square root leads to

$x-{x}_{V}=\pm \sqrt{-\frac{{y}_{V}}{a}}$

and finally to the zeros

${x}_{\mathrm{1,2}}={x}_{V}\pm \sqrt{-\frac{{y}_{V}}{a}}$

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

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