Modeling with Quadratic Equations Flashcards

Study Notes

Profit Model for Soccer Balls

Daily profit for selling soccer balls is modeled by the quadratic equation y = -6x² + 100x - 180.
The graph of this equation decreases at certain intervals due to the nature of the parabola opening downwards.

Vertex of the Parabola

The vertex is at (8.33, 236.67), representing the maximum daily profit of $236.67.
This maximum occurs when soccer balls are priced at $8.33 each.

Zeroes of the Function

Zeroes of the quadratic equation indicate the price points where profit equals zero.
Identifying the zeroes can help determine break-even prices for selling soccer balls.

Achieving Specific Profit Goals

To achieve a daily profit of $150, solve the equation 150 = -6x² + 100x - 180.
The solutions x = 4.53 and x = 12.13 indicate that pricing soccer balls at these values yields the desired profit.

Profit from Football Sales

With soccer balls priced at $7.50, the store earns $232.50 in daily profit.
To meet a total profit goal of $400, additional profit of $167.50 must come from football sales.
The relevant equation for football profit is 167.50 = -4x² + 80x - 150, leading to potential football prices of $5.46 and $14.54.

Intersection of Profit Graphs

Two quadratic equations model profits for soccer balls and footballs respectively.
Intersection points show when the price and profit for each type of ball are equal.
Key intersection points are approximately (8.16, 236.49) and (1.84, -16.49):
- At $8.16, the store makes equal profit (~$236.49) from both balls.
- Pricing at $1.84 results in no profit for either type.

Description

These flashcards focus on understanding quadratic equations in the context of real-world applications, such as profit modeling in a sporting goods store. Each card presents scenarios and questions that test your grasp of the concepts related to quadratic functions.