Modeling with Quadratic Equations
Soccer balls go on sale for $7.50 each. The store also sells footballs and the manager wants to earn a daily profit of $400 from both items. The equation y=-4x2+80x-150 models the store's daily profit, y, for selling footballs at x dollars. Explain how to find the price per football needed to meet the goal.
At $7.50 per soccer ball, they earn a daily profit of $232.50. The store needs to earn a daily profit of $400 - $232.50 = $167.50 from footballs. Solve 167.50 = -4x2 + 80x - 150 to find the price for footballs: x = $5.46 and $14.54.
The table shows some values that satisfy the quadratic equation. Which of the following is a true statement?
B
What is the vertex of the parabola? Round to the nearest hundredth.
B
What are the zeroes of the function? Round to the nearest hundredth.
B and D
A sporting goods store uses quadratic equations to monitor the daily cost and profit for various items it sells. The store's daily profit, y, when soccer balls are sold at x dollars each, is modeled by y=-6x2+100x-180. Why is there an interval over which the graph decreases?
C
What do the zeroes mean in context?
D
The quadratic equation y = -6x2 + 100x - 180 models the store's daily profit, y, for selling soccer balls at x dollars. The quadratic equation y = -4x2 + 80x - 150 models the store's daily profit, y, for selling footballs at x dollars. Use a graphing calculator to find the intersection point(s) of the graphs, and explain what they mean in the context of the problem.
The points of intersection represent when the price and profit are the same for each type of ball. The intersection points are approximately(8.16, 236.49) and (1.84, -16.49). When the store charges $8.16 for each type of ball, they make the same profit from each ball, approximately $236.49. Charging $1.88 provides no profit for either type of ball.
The vertex is located at (8.33, 236.67). What does this point represent in context?
The vertex is the maximum of the graph. The y-coordinate of the maximum is 236.67, which means the greatest possible daily profit from the sale of soccer balls is $236.67. This maximum profit occurs when x = 8.33, or when the store charges $8.33 per soccer ball.
Suppose the store wants to earn a daily profit of $150 from the sale of soccer balls. To earn this profit, what price should the store charge for each soccer ball? Explain how to solve this problem.
You need to solve the equation 150 = -6x2 + 100x - 180. I can subtract 150 from both sides and use the quadratic formula to find x = 4.53 and 12.13. This means that if the store sells soccer balls for $4.53 or $12.13, it will earn a daily profit of $150.