How To Find Vertex From Standard Form

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The Vertex of a Parabola

The vertex of a parabola is the point where the parabola crosses its axis of symmetry.   If the coefficient of the $x^2$ term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the " $U$ "-shape.  If the coefficient of the $x^{\mathrm{two}}$ term is negative, the vertex will exist the highest point on the graph, the point at the top of the " $U$ "-shape.

The standard equation of a parabola is

$y=a{\mathrm{10}}^2+bx+c$ .

But the equation for a parabola tin can also be written in "vertex grade":

$y=a{(\mathrm{ten}-h)}^{\mathrm{two}}+k$

In this equation, the vertex of the parabola is the point $(h,\mathrm{thousand})$ .

You can come across how this relates to the standard equation past multiplying it out:

$\begin{array}{l}y=a(x-h)(x-h)+\mathrm{yard}\\ y=a{x}^2-2ahx+a{h}^2+k\end{array}$ .

This means that in the standard form, $y=a{x}^{\mathrm{ii}}+b\mathrm{ten}+c$ , the expression $-\frac{b}{2a}$ gives the $x$ -coordinate of the vertex.

Example:

Find the vertex of the parabola.

$y=\mathrm{three}{x}^2+12x-12$

Here, $a=3$ and $b=12$ . So, the $x$ -coordinate of the vertex is:

$-\frac{12}{2\left(3\right)}=-2$

Substituting in the original equation to get the $y$ -coordinate, we become:

$\begin{array}{l}y=\mathrm{iii}{\left(-2\right)}^2+12\left(-2\right)-12\\ =-24\end{array}$

And then, the vertex of the parabola is at $(-2,-24)$ .

Source: https://www.varsitytutors.com/hotmath/hotmath_help/topics/vertex-of-a-parabola#:~:text=You%20can%20see%20how%20this,x%20%2Dcoordinate%20of%20the%20vertex.

Posted by: sheppardforgiagether.blogspot.com

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