Modeling with Quadratic Equations

Question 1

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 . Why is there an interval over which the graph decreases?

Revolt · Accepted Answer

If the soccer balls are too expensive, fewer will be sold, reducing profit.

Question 2

Which did you include in your response?

Revolt · Accepted Answer

The vertex is the maximum of the graph.; The maximum daily profit from the sale of soccer balls is $236.67.; The maximum profit occurs when the store charges $8.33 per soccer ball.

Question 3

The table shows some values that satisfy the quadratic equation. Which of the following is a true statement?

Revolt · Accepted Answer

If the store charges $8 per soccer ball, it makes a daily profit of $236.

Question 4

What is the vertex of the parabola? Round to the nearest hundredth.

Revolt · Accepted Answer

(8.33, 236.67)

Question 5

The vertex is located at (8.33, 236.67). What does this point represent in context?

Revolt · Accepted Answer

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.

Question 6

What are the zeroes of the function? Round to the nearest hundredth.

Revolt · Accepted Answer

2.05; 14.61

Question 7

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.

Revolt · Accepted Answer

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.

Question 8

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 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.

Revolt · Accepted Answer

At $7.50 per soccer ball, the profit is $232.50, so the store needs $167.50 in profit from footballs. Set −4x² + 80x − 150 = 167.50 and solve for x. The football price can be about $5.46 or $14.54.

Question 9

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.

Revolt · Accepted Answer

The graphs intersect at approximately (1.84, −16.49) and (8.16, 236.49). These points show prices at which the profit is the same for soccer balls and footballs. At a price of about $8.16, the profit is about $236.49 for each type; at about $1.84, the profit is about −$16.49 for each type.

Question 10

What do the zeroes mean in context?

Revolt · Accepted Answer

If soccer balls are sold for $2.05 or $14.61 each, the store will break even but will not make a profit.

Modeling with Quadratic Equations Answers

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