# How do you find the vertex and intercepts for #x = 1/-32y^2#?

##### 2 Answers

Vertex

The intercepts are only at 1 point, the origin

The axis of symmetry is the x-axis ie

#### Explanation:

This is a quadratic in

So instead of form type

Type 1

Type 2

Instead of the horse shoe type shape being up or down it is left or right.

As the coefficient of

Consider equationtype(1): The

Consider equationtype(2); The

However, in this graph

y-intercept is at

If you write the Parabolic Equation in the form

then the x value for the vertex will be = - b / 2a.

For the x intercept, set y=0, in the equation

For the y intercept, set x=0,

#### Explanation:

If you then plug "- b / 2a " in to the equation you will get the y value and thus the vertex coordinates.

Vertex = ( -b/2a , f ( -b/2a) )

The reason this is true is that the vertex is where the slope = 0 and you can find this by setting the derivative = 0

The derivative is 2ax + b

Setting it = 0 ==> x = -b/2a