# How do you graph and find vertex, directrix axis and vertex for x=-1/12y^2

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By definition, the standard equation of parabola that has a horizontal axis is the following, such that:

`(y - k)^2 = 4p(x - h)`

`(h,k)` represent the coordinates of vertex of parabola

`x = h - p` represents the equation of directrix

`y = k` represents the axis of parabola

You need to convert the given equation to this form, hence, you need to divide the equation by -`1/12` , such that:

`x/(-1/12) = y^2 => y^2 = -12x`

You may identify h,k and p comparing the equations, such that:

`k = 0, h = 0, 4p = -12 => p = -3`

**Hence, evaluating the vertex of parabola yields the origin `(0,0)` , the equation of directrix is `x = 0 - (-3) = 3` and the axis of parabola is the y axis.**

`x=(-1/12)y^2`

`y^2=-12x=4(-3)x`

compare it with standard equation of parabola

`y^2=4ax`

So vertex =(0,0)

directrix

x=-(-3)

x=3

axis of parabola

y=0