Accepted Answer

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Tickets= -0.2x^2+12x+11

a. Does the graph of this equation open up or down? How did you determine this?

It opens down, since the coefficient of the x^2 term is negative.

b. Describe what happens to the tickets sales as time passes?

First, the ticket sales rise. Then they reach a peak, and fall back to zero.

c. Use the quadratic equation to determine the last day that tickets will be sold. (Note: Write your answer in terms of the number of days after ticket sales begin.)

Set it to 0:
0 = -0.2x^2+12x+11

Multiply by -5:

x^2 - 60x - 55 = 0

Quadratic formula:

x = (-b +/- sqrt(b^2 - 4ac))/(2a)

x = (60 +/- sqrt(60^2 - 4*-55))/2

x = 60.903 or -0.903

Negative doesn't make sense, so:

x = 60.903 days (about 61 days, or 60 days after x = 1)

The last day with sales will be around day 60.

d. Will tickets peak or be at a low during the middle of the sale? How do you know?

They will be at a peak during the middle of the sale. Since the graph opens downwards, the vertex will be at the maximum of the function.

e. After how many days will the peak or low occur?

-b/2a

-12/(2*-0.2)

x = 30 days (29 days after x = 1)

f. How many tickets will be sold on the day when the peak or low occurs?

Plug in x = 30:

-0.2*30^2 + 12*30 + 11

= 191 tickets

g. What is the point of the vertex? How does this number relate to your answers in parts e and f?

The vertex is at:
(30,191)
The x coordinate is the day that the most were sold. The y coordinate is the number of tickets.

h. How many solutions are there to the equation ? How do you know?

There are two solutions to the equation, because b^2-4ac is positive.
12^2 - 4*-0.2*11 = 152.8

i. What do the solutions represent? Is there a solution that does not make sense? If so, in what ways does the solution not make sense?

The solutions are the days when 0 tickets are sold. The negative solution (-0.903) doesn't make sense, since that time would be before the tickets even went on sale in the first place.

Let me know if you have any questions before you press "Accept". Thanks,
Scott

Expert:	Scott
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Answered:	9/10/2008

MIT Graduate

College degree in math... proficient in all levels -- from algebra to calculus