Modeling with Quadratic Equations Flashcards
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Modeling with Quadratic Equations Flashcards

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@TenaciousFeynman9892

Questions and Answers

Why is there an interval over which the graph decreases?

C

The table shows some values that satisfy the quadratic equation. Is this statement true or false?

True

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

B

What does the vertex (8.33, 236.67) represent in context?

<p>The vertex is the maximum of the graph. The y-coordinate of the maximum is 236.67, which reflects the greatest possible daily profit from the sale of soccer balls.</p> Signup and view all the answers

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

<p>B and D</p> Signup and view all the answers

What do the zeroes mean in context?

<p>D</p> Signup and view all the answers

To earn a daily profit of $150 from the sale of soccer balls, what price should the store charge for each soccer ball?

<p>x = 4.53 and 12.13</p> Signup and view all the answers

How to find the price per football needed to meet the profit goal of $400?

<p>x = 5.46 and 14.54</p> Signup and view all the answers

What do the intersection points of the graphs represent?

<p>The points of intersection represent when the price and profit are the same for each type of ball.</p> Signup and view all the answers

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.

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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.

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