Question 199430
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Can't help you unless you share the actual function, except in very general terms.


If the coefficient on the *[tex \Large x^2] term is positive, then the curve opens upward, otherwise it opens down.  From the way part c is worded, I suspect it opens down -- otherwise, there wouldn't be a "last day"


With that presumption, the ticket sales will increase to a peak and then fall off (part b)


To answer part c, set the function equal to zero and then solve the quadratic.  You would exclude any zero or negative root because the last day can't be the first day and the last day can't happen before the first day.


part d:  see the answer to part b.  


Given a quadratic function *[tex \Large f(x) = ax^2 + bx + c], the vertex occurs at the x-coordinate *[tex \Large \frac{-b}{2a}].  That will answer part e.


Evaluate the function at the answer to part e, that is substitute that value for x in the function and calculate the value of the function to get part f.


The answers to e and f are the coordinates of the ordered pair representing the vertex.


for part h, I can't exactly say without looking at the actual function, however using the assumption I used for part c, the vertex is probably above the x-axis, hence the curve intersects the x-axis in two places -- hence two solutions.


Each solution represents a day when ticket sales are zero.  A negative solution would not make sense in terms of what is going on.  Of course ticket sales are zero at any time before you start selling them -- but knowing that isn't very helpful when trying to model the situation.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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