1. Suppose you are an event coordinator for a large performance theater. One of the hottest new Broadway musicals has started to tour and your city is the first stop on the tour. You need to supply information about projected ticket sales to the box office manager. The box office manager uses this information to anticipate staffing needs until the tickets sell out. You provide the manager 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 this equation open up or down? How did you determine this?
b. Describe what happens to the tickets sales as time passes.
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.
d. Will tickets peak or be at a low during the middle of the sale? How do you know?
e. After how many days will the peak or low occur?
f. How many tickets will be sold on the day when the peak or low occurs?
g. What is the point of the vertex? How does this number relate to your answers in parts e. and f?
h. How many solutions are there to the equation -0.2x^2+12x+11=0 ? How do you know?
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?
a. To determine whether the graph of the equation opens up or down, we can look at the coefficient of the quadratic term, which is -0.2x^2. Since the coefficient is negative (-0.2), the graph will open down.
b. As time passes, the ticket sales will go through a peak and then decrease. Initially, ticket sales will start slowly but then pick up speed until they reach the peak. After the peak, ticket sales will gradually decline.
c. To determine the last day that tickets will be sold, we need to find the x-value where the ticket sales equation equals zero. We set the equation, -0.2x^2 + 12x + 11 = 0, and solve for x. The resulting value of x will represent the last day that tickets will be sold.
d. Tickets will be at a low during the middle of the sale because the graph opens down, indicating a decrease in ticket sales. Since the coefficient of the quadratic term is negative, the graph forms a "U" shape with a peak.
e. To find out after how many days the peak or low occurs, we can find the x-value of the vertex of the quadratic equation. The x-value of the vertex represents the midpoint of the parabola. We can determine this by finding the axis of symmetry, which is given by the formula x = -b / (2a), where a and b are the coefficients of the quadratic equation. In this case, a = -0.2 and b = 12.
f. Once we determine the day when the peak or low occurs (as mentioned in part e), we can substitute that day's value of x into the quadratic equation to find out how many tickets will be sold on that day.
g. The point of the vertex is the lowest or highest point of the parabola depending on whether it opens up or down. In this case, since the graph opens down, the vertex will be the highest point. The vertex represents the maximum number of ticket sales during the entire period, and the y-coordinate of the vertex will give us that maximum value. This number relates to the answers in parts e and f because it tells us the number of tickets sold on the day of the peak or low.
h. To determine the number of solutions to the equation -0.2x^2 + 12x + 11 = 0, we can use the discriminant (b^2 - 4ac) of the quadratic equation, where a = -0.2, b = 12, and c = 11. If the discriminant is positive, there are two distinct solutions. If it's zero, there is one repeated solution. If it's negative, there are no real solutions. By calculating the discriminant, we can determine the number of solutions.
i. The solutions to the equation represent the real values of x at which the ticket sales equation equals zero. These solutions correspond to the days when the tickets will stop being sold. Sometimes, there may be a solution that does not make sense, such as a negative value for x, which would mean that the tickets stopped being sold before they even went on sale. In such cases, that solution is considered extraneous and does not make sense in the context of the problem.