Question

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

Answer 1

Vertex`->(x,y)=(0,0)`

The intercepts are only at 1 point, the origin

The axis of symmetry is the x-axis ie `y=0`

Explanation:

This is a quadratic in `y` instead of `x`

So instead of form type `y=ax^2+bx+c ....EquationType(1)`
`" we have: " x=ay^2+by+c...EquationType(2)`

Type 1`->nn" or "uu`
Type 2`-> sub" or "sup`

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

As the coefficient of `y^2` is negative the general shape is `sup`

Consider equationtype(1): The `b` from `bx` moves the graph left or right in that `x_("vertex")=(-1/2)xxb`

Consider equationtype(2); The `b` from `by` moves the graph up or down in that `y_("vertex")=(-1/2)xxb`

However, in this graph `b=0` so the axis of symmetry coincides with the x-axis.

y-intercept is at `x=0`

`=>0=-1/32 y^2" "=>" "y=0` which is only 1 value (single point) so the graph does not `ul("cross")` the y-axis but the axis is tangential to the vertex.

Answer 2

If you write the Parabolic Equation in the form `y = ax^2 + bx + c`
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