AnswersFL-1200330-Algebra 2 CRModeling with Quadratic Equations

Modeling with Quadratic Equations Answers

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1
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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?

Question illustration
A
If the store sells more soccer balls, they can decrease the price.
B
If the soccer balls are returned for a refund, the store will lose money.
C
If the soccer balls are too expensive, fewer will be sold, reducing profit.
2
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The table shows some values that satisfy the quadratic equation. Which of the following is a true statement?

A
If the store charges $5 per soccer ball, it can expect to sell 170 of them.
B
If the store charges $8 per soccer ball, it makes a daily profit of $236.
C
If the store sells 12 soccer balls, it makes $156 in profit.
D
The greater the selling price, the greater the profit.
3

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

A
(16.67, 1,306.67)
B
(8.33, 236.67)
C
(0.03, –177)
D
(8.33,0)
E
(2, 0)
F
(14.61, 0)
4

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

Answer:

The vertex represents the maximum point of the quadratic function. In context, the x-coordinate (8.33) represents the price per soccer ball that will result in the maximum profit, and the y-coordinate (236.67) represents the maximum daily profit of $236.67.

5

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

A
0
B
2.05
C
8.33
D
14.61
6

What do the zeroes mean in context?

A
If soccer balls are sold for $2.05 each, the store will make a daily profit of $14.61.
B
If soccer balls are sold for $14.61 each, the store will make a daily profit of $2.05.
C
If soccer balls are sold for $2.05 or $14.61 each, the store will maximize their daily profit.
D
If soccer balls are sold for $2.05 or $14.61 each, the store will break even but will not make a profit.
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.

Answer:

To find the price that results in a daily profit of $150, set the profit equation equal to 150: 150 = -6x² + 100x - 180. Subtract 150 from both sides to get -6x² + 100x - 330 = 0. Using the quadratic formula, the solutions for x (the price) are approximately $4.53 and $12.13. Therefore, the store should charge either $4.53 or $12.13 per soccer ball to earn a $150 profit.

8

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?

Answer not available
10

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

A
If the store charges $5 per soccer ball, it can expect to sell 170 of them.
B
If the store charges $8 per soccer ball, it makes a daily profit of $236.
C
If the store sells 12 soccer balls, it makes $156 in profit.
D
The greater the selling price, the greater the profit.
12

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

A
(16.67, 1,306.67)
B
(8.33, 236.67)
C
(0.03, –177)
D
(8.33,0)
E
(2, 0)
F
(14.61, 0)
14

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

A
0
B
2.05
C
8.33
D
14.61
15

What do the zeroes mean in context?

A
If soccer balls are sold for $2.05 each, the store will make a daily profit of $14.61.
B
If soccer balls are sold for $14.61 each, the store will make a daily profit of $2.05.
C
If soccer balls are sold for $2.05 or $14.61 each, the store will maximize their daily profit.
D
If soccer balls are sold for $2.05 or $14.61 each, the store will break even but will not make a profit.
17

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.

Question illustration
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.

18

Question text not available

Answer not available
19

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.

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.

20

Which did you include in your response?

A
points of intersection represent when the price and profit are the same for each type of ball.
B
The intersection points are approximately(8.16, 236.49) and (1.84, –16.49).
C
When the store charges $8.16 for each type of ball, they make the same profit from each ball, approximately $236.49.
D
Charging $1.88 provides no profit for either type of ball.

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