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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

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

square root leads to

$x-{x}_{S}=\pm \sqrt{-{y}_{S}}$

and finally to the zeros

${x}_{\mathrm{1,2}}={x}_{S}\pm \sqrt{-{y}_{S}}$