Find the standard form of the equation of the parabola with its vertex at the origin and with point (3,6).

Question

whpalmer4 · Answer

Vertex form is $$y=a(x-h)^2+k$$with vertex at $h,k$ and $a$ a constant you set to make the parabola fit your point.  Then rewrite in standard form which is $$y=ax^2+bx+c$$
(Note that a,b,c  are not necessarily the same, I'm just showing the general form!)

anonymous · Answer

I need it in standard form like x^2=4py :/

whpalmer4 · Answer

So rearrange it into that form!

anonymous · Answer

There is a in one equation and p in another though

whpalmer4 · Answer

It all turns into numbers to fit the parabola.

whpalmer4 · Answer

Where is your vertex?

anonymous · Answer

Vertex is 0,0

whpalmer4 · Answer

Okay, h=0,k=0, plug those into my first formula, what do you get?

anonymous · Answer

y=a(x−h)2+k
6 =a(3-0)^2+0
6= 9a
a= 2/3

whpalmer4 · Answer

So what is your final equation?

anonymous · Answer

y=2/3x^2

whpalmer4 · Answer

And with x^2 all alone on one side, that is?

anonymous · Answer

x^2 = 3/2y
OMG THANKS!! :)

whpalmer4 · Answer

Best to write that as (2/3)x^2, btw, just so there is no doubt about the meaning

whpalmer4 · Answer

Yeah, and my formula will work even if the vertex is elsewhere...try it and see :-)

anonymous · Answer

do you mind helping me with another question?

whpalmer4 · Answer

Sorry, I have to leave. Post and tag me, and I'll have a look when I return. Hopefully someone will help you before that.