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Old 10-14-2008, 07:36 PM   #1
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Default vertex of a parabola

can someone tell me how to do this with a trinomial, the trinomial is x(2)+8x-10

please dont rag on the easy math i just cant remember the formula
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Old 10-14-2008, 08:04 PM   #2
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Default Re: vertex of a parabola

It's -b/2a for the x coordinate. Then plug that x coordinate into the equation for x and find the y coordinate.
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Old 10-14-2008, 08:15 PM   #3
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Default Re: vertex of a parabola

It's a lot easier with derivatives. In your case let's see: The derivative is 2x + 8. Set that equal to 0. x = -4. The vertex will be at (-4, -26)

What troy said is also correct and probably used more often in lower math classes. The only problem is that only holds true to the format ax^2 + bx + c. If you want a quick lesson on how to find derivatives of polynomials give me a PM. It's really simple that you'll be like "Holy ****! My dog can do that!". Once you have the derivative you set it equal to 0 and it will give you x coordinates to all relative vertexes.

Or...using a graphing calculator: Graph the polynomial then hit 2nd, trace (will go to calc) and scroll down to maximum/minimum. It will ask for left and right bound then a guess. On left bound, you select a point left of the peak. Right bound, same thing only to the right. Then you guess any point in between, doesn't matter. It will then give you approximations to the actual point if not the exact.
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Old 10-14-2008, 08:30 PM   #4
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Default Re: vertex of a parabola

I hate to be that dude, but. . .
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Old 10-14-2008, 08:33 PM   #5
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Default Re: vertex of a parabola

Quote:
Originally Posted by Illmatic View Post
I hate to be that dude, but. . .
Nice find, but. . .

I tend to ask help for math questions like this rather than google it because I tend to find a real person's explanation to be easier to understand than a PhD. Just my experiences though.
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