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The answer depends on the form in which the quadratic function is given.
If it is y = ax2 + bx + c
then the x-coordinate of the turning point is -b/(2a)
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TaigaOn a graph the turning point is the bottom of of the 'bowl' for y = x^(2) + Bx + C
If the graph is of the form y = -x^(2) + Bx + C , then it is 'umbrella' shaped, and the turning point is the top of curve.
Mathemtically for graphs of the form y = ax^(2) + Bx + C
We differentiate and equate to zero.
dy/dx = 2ax +B = 0
Hence the x-coordinate of the turning point is x = -B/2A
This value of 'x' can be substituted into the original ( undifferentiated) equation to find the y-coordinate.
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