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omg: determine the point on the curve where the tangent line to the curve is horizontal. (The answer must be a POINT on the curve.)
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tangent line horizontal at the vertex, so you are really being asked to find the vertex is \[y=-x^2-6x+2\]
oh i know find derivative! right?
first coordinate of the vertex is always \(-\frac{b}{2a}\) which in your case is \(-\frac{-6}{2\times -1}=-3\)
and then plug in what x is to the normal function.
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yes i got that :)
thank you @satellite73 you refreshed my memory =]
or you can take the derivative, set it equal to zero and solve, which is rather silly because you will still get \(-\frac{b}{2a}\) no need for calculus when you have a quadratic
yw
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