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Mathematics 20 Online
OpenStudy (anonymous):

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.)

OpenStudy (anonymous):

OpenStudy (anonymous):

tangent line horizontal at the vertex, so you are really being asked to find the vertex is \[y=-x^2-6x+2\]

OpenStudy (anonymous):

oh i know find derivative! right?

OpenStudy (anonymous):

first coordinate of the vertex is always \(-\frac{b}{2a}\) which in your case is \(-\frac{-6}{2\times -1}=-3\)

OpenStudy (anonymous):

and then plug in what x is to the normal function.

OpenStudy (anonymous):

yes i got that :)

OpenStudy (anonymous):

thank you @satellite73 you refreshed my memory =]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yw

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