You can put this solution on YOUR website!
Well, I'm sorry to hear that you're having trouble with your appendix but you really should see a doctor, it might be appendicitis.
But back to the problem:
Given the equation for expected ticket sales (T):
a) The graph opens up. You know (or should know) this because the coefficient of the
term is positive (+1). If it were negative, the graph would open down.
b) The number of tickets, T, first goes down until day 3 (x = 3), then it goes up.
c) There's no way to answer this question without first knowing how many tickets were for sale to begin with. This information was not provided.
d) Since the graph opens up, the vertex is a minimum and the tickets will be at a low, but whether this is at the middle of the sale cannot be determined since you have not provided the total number of tickets for sale.
e) The minimum (low) occurs at 3 days (x=3).
f) The number of tickets, T, sold at day 3 is 7 (T = 7 when x = 3).
g) The point of the vertex is (3, 7) and this is determined either from the graph or algebraically:
Find the x-coordinate by:
Now plug this into the quadratic eqation to find y (or T in this problem)
The vertex is at (3, 7)
h) There are two solutions to the given quadratic equation. The number of solutions is equal to the highest power of the independent variable (x) which is 2 in this problem.
i) The solutions to this equation are not real numbers so they are meaningless.
The solutions are meaningless because the graph never intersects the x-axis indicating that the number of tickets sold never reaches zero.
In the graph, T (Expected number of tickets sold)is the vertical axis and x (Days after start of sale) is the horizontal axis.