SOLUTION: The word problem reads as follows: An event coordinator for a theater has requested information about projected ticket sales. you provide him with a quadratic equation that models

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Question 96325: The word problem reads as follows: An event coordinator for a theater has requested information about projected ticket sales. you provide him with a quadratic equation that models the expected number of ticket sales for each day x. (x=1 is the day tickets go on sale). Tickets =-0.2x^2+12x+11.
A. does the graph of the equation open or down and how this is determined?
b. What happens to ticket sales as time passes?
c. use the equation to determine the last day that tickets will be sold.
I am very confused on how to work this problem out. Thank you for nay help you can provide.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The word problem reads as follows: An event coordinator for a theater has requested information about projected ticket sales. you provide him with a quadratic equation that models the expected number of ticket sales for each day x. (x=1 is the day tickets go on sale). Tickets = -0.2x^2+12x+11.
:
A. does the graph of the equation open up or down and how this is determined?
:
It opens downward. This can be determined by the fact that the coefficient of the squared term (-2x^2 in this equation) in a quadratic equation, is negative. It also means that it has a maximum.
If it were positive, it would open up-ward and have a minimum
:
:
b. What happens to ticket sales as time passes?
:
Drawing the actual graph of this equation will make it easy to understand:
+graph%28+300%2C+200%2C+-10%2C+80%2C+-20%2C+200%2C+-.2x%5E2%2B12x%2B11%29+
:
Since it has a maximum, You can see ticket sales increase to a maximum
at about day 30 then decrease to 0 on about day 60
:
:
c. use the equation to determine the last day that tickets will be sold.
You can determine this without the graph by make the equation = 0 and solving
for x:
-.2x^2 + 12x + 11 = 0
:
The quadratic formula is needed here: a= -.2; b = 12; c = 11
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
x+=+%28-12+%2B-+sqrt%28+12%5E2+-+4+%2A+-.2+%2A+11+%29%29%2F%282%2A-.2%29+
:
x+=+%28-12+%2B-+sqrt%28+144+-+%28-8.8%29+%29%29%2F%28-.4%29+
:
x+=+%28-12+%2B-+sqrt%28+144+%2B+8.8+%29%29%2F%28-.4%29+
:
x+=+%28-12+%2B-+sqrt%28+152.8%29%29%2F%28-.4%29+
:
x+=+%28-12+-+12.36%29%2F-.4; we only want the positive solution here
:
x+=+%28-24.36%29%2F-.4
:
x = +60.9 or on the 61st day, no tickets sold, so day 60 would be the last day
that tickets are sold, note that is what the graph indicates.
:
In fact if you substitute 60 for x in the graph, you will see that 11 tickets
were sold on the last day
:
You can plot this graph yourself using this table
x | y
-------
-1 |neg
0 | 11
10 |111
20 |171
30 |191
40 |171
50 |111
60 | 11
70 | neg
:
That should help you see what's going on here, Any question?