Parabola someone please help me

Find the vertex, the axis of symmetry, the focus and the directrix of each parabola.

y^2=12x

y^2 = 12x
(y - k)^2 = 4p (x - h), where the focus is (h + p, k) and the directrix is x = h - p
vertex (h , k) = (0 , 0)
the parabola opens right,
axis of symmetry: y = 0
4p = 12
p = 3
focus(3 , 0)
directrix=> x = -3

If you interchange x and y we get y = x^2/12

Use your textbook's procedure to find the parts you wish for this function.

For each aspect, just exchange the x and y characteristics to identify the needs for the original function.

For example: for y = x^2/12, axis of symmetry is x = 0, so the axis of symmetry for the original function is y = 0

And so on!

The graph of the original function is a parabola that opens to the right with vertex at (0, 0).

If you know the quadratic formula, you know the formula for the x coordinate of the vertex, and also the line of symmetry.

The stuff outside the radical (-b/2a) is that x coordinate.

Here, rather than being a parabola that opens up or down, this one opens to the side.
That's what happens when the y is squared instead of the x.

Then if we solve for x, this is x = 1/12 * y^2

So a = 1/12 and b = c = 0

Then -b/2a = 0/2a = 0

Normally this would be the x coordinate, but remember, this is one where the y coordinate is squared, so rather than this being x = 0,
it will be y = 0 instead.

So the axis of symmetry is y = 0, or the x axis.

To find the actual point, sub in y = 0 into the equation and solve for x.

You should find that the vertex is at the origin, (0,0)

The focus must be on the axis of symmetry, so the y coordinate must be 0

I never remember this formula, so I'll just give it to you

The x coordinate is [1-(b^2 - 4ac)]/4a

Here, 1/4a = 12/4 = 3