Question

Therefore, the vertex of the problem I previously asked would be 1,0. Is that right?

Answer 1

Well, that was a search problem for me.
here is what drwls told you:
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Either -x^2 + 2x + 1 = 0 or
x^2 -2x -1 = 0 are in "proper quadratic form".
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you are evidently doing the parabola
y = -x^2 + 2 x + 1
now if you are looking for the x axis intercepts
0 = [-2 +/- sqrt (4 + 4)] /-2
which is
+1 +/- sqrt 2
since y gets big negative when x gets big + or -, it opens down (sheds water)
the vertex is where x = 1 since the intercepts are 1 + sqrt 2 and 1 - sqrt 2
when x = 1, y = 2
so the vertex is at (1,2)

Answer 2

Thank you.

Answer 3

Sharon
looking back, you gave people several different versions of this, with different signs for different terms. That is why you have a whole mess of replies.

Answer 4

I'm not sure what you mean because I thought I was clear with my question. But thank you for your help anyway.