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

Mar 16, 2017

Vertex$\to \left(x , y\right) = \left(0 , 0\right)$

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 $y = a {x}^{2} + b x + c \ldots . E q u a t i o n T y p e \left(1\right)$
$\text{ we have: } x = a {y}^{2} + b y + c \ldots E q u a t i o n T y p e \left(2\right)$

Type 1$\to \cap \text{ or } \cup$
Type 2$\to \subset \text{ or } \supset$

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 $\supset$

Consider equationtype(1): The $b$ from $b x$ moves the graph left or right in that ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times b$

Consider equationtype(2); The $b$ from $b y$ moves the graph up or down in that ${y}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times b$

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

y-intercept is at $x = 0$

$\implies 0 = - \frac{1}{32} {y}^{2} \text{ "=>" } y = 0$ which is only 1 value (single point) so the graph does not $\underline{\text{cross}}$ the y-axis but the axis is tangential to the vertex.

Jul 25, 2017

If you write the Parabolic Equation in the form $y = a {x}^{2} + b x + c$
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