Therefore, the vertex of the problem I previously asked would be 1,0. Is that right?
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)
Thank you.
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.
I'm not sure what you mean because I thought I was clear with my question. But thank you for your help anyway.
To determine the vertex of a quadratic problem, you need to have the equation or key information about the problem.
If you have the equation in the standard form "y = ax^2 + bx + c", the x-coordinate of the vertex, denoted as "h," can be found using the formula "h = -b / (2a)". Then, to find the y-coordinate of the vertex, substitute the value of "h" into the equation.
If you have the quadratic equation in the vertex form "y = a(x - h)^2 + k", the vertex is represented by the point (h, k).
It seems that you mentioned the vertex of "1,0" without providing any equation or further context. Without additional information, it is not possible to determine if the vertex is indeed located at (1, 0). Please provide the equation or more details to accurately solve for the vertex.