Question

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

Answer 1

Refer to the explanation.

Explanation:

The standard form of a parabola is `y=ax^2++bx+c`, where `a!=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=(-b)/(2a)`

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

`y=a((-b)/(2a))^2+b((-b)/(2a))+c`

Example:

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

Step 1. Find the x-coordinate of the vertex

`x=(-4)/(2*1)`

`x=-4/2`

`x=-2` `larr` x-coordinate of the vertex

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

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

`y=(-2)^2+4(-2)-9`

`y=4-8-9`

`y=-13` `larr` y-coordinate of the vertex

The vertex is `(-2,-13)`.

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

Answer 2

See the wholesome answer, in the explanation.

Explanation:

When the axis and the perpendicular tangent at the vertex

`V ( alpha, beta )`

are parallel to the coordinate axes, the standard form is

`( y - beta )^2 = +- 4a ( x - alpha)` or

`( x - alpha )^2 = +- 4a ( y - beta )`

Graph for

`(y+3)^2 = - 4 ( x - 1 )`, and vice versa, `( x - 1 )^2 = - 4 ( y + 3)`:
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

`ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0` represents a parabola,

and my standard form here is

`((y-beta) - m( x - alpha ) )^2 = +-4a(m ( y-beta ) +( x - alpha ) )` or

`+- 4a ((y-beta) - m( x - alpha ) ) = ( m ( y-beta ) +( x - alpha ) )^2`

Example: `( ( y - 2 ) - 3 ( x + 2 ))^2= -2 ( 3 ( y -2 ) + ( x + 2 ))`

Vertex : `V ( alpha, beta ) = ( - 2, 2 )`

Axis: `( y - 2 ) - 3 ( x + 2 ) = 0`

Tangent at V: `3 ( y -2 ) + ( x + 2 ) = 0`

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