Question 602595
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^2\ +\ 36x]


Graph is a parabola opening upward.


Vertex is at 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v = \frac{-b}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)]


So fill in the numbers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v = \frac{-36}{2(1)}]


And do the arithmetic to find *[tex \LARGE x_v], then plug that value into


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)\ =\ (x_v)^2\ +\ 36(x_v)]


to find the value of the function at the vertex which is the minimum value of the function and hence the smallest product.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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