# How do you find the vertex of a parabola in standard form?

Aug 11, 2018

Refer to the explanation.

#### Explanation:

The standard form of a parabola is $y = a {x}^{2} + + b x + c$, where $a \ne 0$.

The vertex is the minimum or maximum point of a parabola. If $a > 0$, the vertex is the minimum point and the parabola opens upward. If $a < 0$, the vertex is the maximum point and the parabola opens downward.

To find the vertex, you need to find the x- and y-coordinates.

The formula for the axis of symmetry and the x-coordinate of the vertex is:

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

To find the y-coordinate of the vertex, substitute the value for $x$ into the equation and solve for $y$.

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

Example:

Find the vertex of $y = {x}^{2} + 4 x - 9$, where: $a = 1$, $b = 4$, and $c = - 9$.

Step 1. Find the x-coordinate of the vertex

$x = \frac{- 4}{2 \cdot 1}$

$x = - \frac{4}{2}$

$x = - 2$ $\leftarrow$ x-coordinate of the vertex

Step 2. Find the y-coordinate of the vertex.

Substitute $- 2$ for $x$ and solve for $y$.

$y = {\left(- 2\right)}^{2} + 4 \left(- 2\right) - 9$

$y = 4 - 8 - 9$

$y = - 13$ $\leftarrow$ y-coordinate of the vertex

The vertex is $\left(- 2 , - 13\right)$.

graph{y=x^2+4x-9 [-9.71, 10.29, -13.68, -3.68]}

Aug 11, 2018

See the wholesome answer, in the explanation.

#### Explanation:

When the axis and the perpendicular tangent at the vertex

$V \left(\alpha , \beta\right)$

are parallel to the coordinate axes, the standard form is

${\left(y - \beta\right)}^{2} = \pm 4 a \left(x - \alpha\right)$ or

${\left(x - \alpha\right)}^{2} = \pm 4 a \left(y - \beta\right)$

Graph for

${\left(y + 3\right)}^{2} = - 4 \left(x - 1\right)$, and vice versa, ${\left(x - 1\right)}^{2} = - 4 \left(y + 3\right)$:
graph{((y+3)^2 + 4 ( x - 1 ))(( x - 1 )^2 + 4 ( y + 3 ))((x-1)^2+(y+3)^2-0.01)=0[-5 7 -6 0]}

If ab = h^2, the general 2nd-degree equation

$a {x}^{2} + 2 h x y + b {y}^{2} + 2 g x + 2 f y + c = 0$ represents a parabola,

and my standard form here is

${\left(\left(y - \beta\right) - m \left(x - \alpha\right)\right)}^{2} = \pm 4 a \left(m \left(y - \beta\right) + \left(x - \alpha\right)\right)$ or

$\pm 4 a \left(\left(y - \beta\right) - m \left(x - \alpha\right)\right) = {\left(m \left(y - \beta\right) + \left(x - \alpha\right)\right)}^{2}$

Example: ${\left(\left(y - 2\right) - 3 \left(x + 2\right)\right)}^{2} = - 2 \left(3 \left(y - 2\right) + \left(x + 2\right)\right)$

Vertex : $V \left(\alpha , \beta\right) = \left(- 2 , 2\right)$

Axis: $\left(y - 2\right) - 3 \left(x + 2\right) = 0$

Tangent at V: $3 \left(y - 2\right) + \left(x + 2\right) = 0$

Graph, with the axis and the tangent at $V \left(- 2 , 2\right)$:
graph{(( ( y - 2 ) - 3 ( x + 2 ))^2 + 2 ( 3 ( y -2 ) + ( x + 2 )))(( y - 2 ) - 3 ( x + 2 ))( 3 ( y -2 ) + ( x + 2 )) = 0}