How do you find the vertex and intercepts for x = 1/-32y^2x=132y2?

2 Answers
Mar 16, 2017

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

The intercepts are only at 1 point, the origin

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

Explanation:

This is a quadratic in yy instead of xx

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.

Tony BTony B

Jul 25, 2017

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