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View Full Version : Find a vertex of parabola with only yint and a point.


nmstar56
Dec 9, 2009, 08:54 PM
I'm given the vertex: (x,6.17)
y-Int: (0,5)
One x-int: (10.42, 0)

How do I find the rest of the vertex, and then find the equation for the parabola.

galactus
Dec 10, 2009, 11:22 AM
I'm given the vertex: (x,6.17)
y-Int: (0,5)
One x-int: (10.42, 0)

How do I find the rest of the vertex, and then find the equation for the parabola.

There are various ways to appproach this, but you could use

y=a(x-h)^{2}+k

Where h and k are the vertex coordinates. Plug in your knowns and solve for h and a.

0=a(10.42-h)^{2}+6.17

5=a(0-h)^{2}+6.17

Solve for a and h.

galactus
Dec 15, 2009, 06:56 AM
Thanks, Unknown. You thought it was nice, but the poster must've have thought so because they never responded back.

Unknown008
Dec 15, 2009, 07:20 AM
That's how most of them are, look for an answer, never to look if they got an answer or not. I'm sure most of them think they'll get the answer a second after they post their question :rolleyes: