Using a quadratic equation by using graph in finding the price, net price, earnings 

To solve problems involving a quadratic equation using a graph, you can typically follow these steps:

1. **Set Up the Quadratic Equation**: Usually, the quadratic equation represents some form of profit, revenue, or cost function, often of the form $ P(x) = ax^2 + bx + c $, where $ P(x) $ could represent profit, revenue, or cost depending on the context, and $ x $ represents the quantity or price.

2. **Graph the Quadratic Equation**: Plot the quadratic equation on a graph. The graph will typically be a parabola, which opens upwards if $ a > 0 $ and downwards if $ a < 0 $.

3. **Identify Key Points on the Graph**:
    - **Vertex**: The vertex of the parabola gives you either the maximum or minimum value. For a profit function, the vertex will give you the maximum profit and the corresponding price or quantity.
    - **X-Intercepts**: These points represent where the profit, revenue, or cost equals zero. They are the break-even points or the price/quantity where no profit is made.

4. **Find the Price/Quantity for Maximum or Minimum**:
    - The vertex $ x = -\frac{b}{2a} $ gives the price or quantity at which the maximum or minimum occurs.
    - Substitute this value back into the equation to find the corresponding net price or earnings.

5. **Analyze the Results**: 
   - Use the graph to interpret the solution, identifying the price at which the maximum profit occurs and the earnings at that price.

### Example Problem:

Suppose the profit $ P(x) $ from selling $ x $ units of a product is given by the quadratic equation $ P(x) = -2x^2 + 12x - 16 $.

1. **Graph the Function**: The graph of $ P(x) = -2x^2 + 12x - 16 $ is a downward-opening parabola because the coefficient of $ x^2 $ is negative.

2. **Vertex (Maximum Profit)**: The vertex $ x = -\frac{b}{2a} $:
   \[
   x = -\frac{12}{2(-2)} = 3
   \]
   So, the maximum profit occurs when $ x = 3 $.

3. **Net Price/Earnings at Maximum Profit**:
   Substitute $ x = 3 $ into $ P(x) $ to find the earnings:
   \[
   P(3) = -2(3)^2 + 12(3) - 16 = -18 + 36 - 16 = 2
   \]
   So, the maximum profit is $2.

4. **X-Intercepts (Break-Even Points)**:
   Solve $ -2x^2 + 12x - 16 = 0 $ using the quadratic formula to find the break-even points.

### Questions for Further Exploration:

1. How does the value of $ a $ affect the shape of the parabola?
2. How can you determine if a quadratic equation will produce a maximum or a minimum value?
3. What do the roots of the quadratic equation tell you in a business context?
4. How would changing the linear term $ b $ affect the profit function?
5. What practical limitations might there be when interpreting a quadratic model in real-world scenarios?

### Tip:
When graphing a quadratic equation, always pay attention to the direction in which the parabola opens, as it indicates whether you're dealing with a maximum or minimum situation.

Using a quadratic equation by using graph in finding the price, net price, earnings

Learn how to use quadratic equations and graphs to find key business metrics like price, net price, and earnings. This tutorial includes an example with step-by-step calculations using the quadratic equation P(x) = -2x^2 + 12x - 16, ideal for students in Grades 9-12.

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