Albebra
posted by Joel
Find the vertex of the function y = –x2 + 2x + 8.
It is where the derivative 2x + 2 = 0.
Since you probably haven't learned about derivatives, do it this way:
Rewrite is as
x^2 +2x +8 = (x1)^2 + 7
Now look at the form of the formula on the right. It can never be greater than 7. It will have its highest value when (x1) = 0
it shows me different points as the answer. Im not sure how to find that point. Here are the points: (1,7) or (1,9) or (4,2) or (1,9)
Looks like we had the same idea too! :)
We're working together...He's right those are the pints how do we find which one it is?
Here's a quick way to find the vertex.
Use the format: [b/2a, (4acb^2)/4a]
a = 1, b = 2, and c = 8
Substitute the values to find (x, y).
This will be your vertex.
Did you not see Drwls' second solution?
If you don't recognize the method of "completing the square", which he used, it would be hard to tell what grade level you are in.
Another "quick" way to find the vertex of any parabola in the form y=ax^2+bx+c is this:
the x value of the vertex is =b/(2a)
substitute that back in the equation to get the y value.
Hi MathGuru, looks like we were seconds apart, lol
I also messed up by missing the negative sign in front of the b
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