Quadratic Modeling

Grade 10 · mathematics · 98 practice problems · read aloud

🔊 Listen to this explanation

Quadratic Modeling: The Big Picture

Quadratic modeling uses quadratic functions (equations of the form y = ax² + bx + c) to represent real-world situations. The graph is a smooth, U-shaped curve called a parabola. This is incredibly useful for predicting outcomes involving area, projectile motion, and profit maximization. 🚀

Step-by-Step Problem Solving

  1. Identify the Scenario: Look for keywords like "area," "height of a thrown ball," or "maximum profit."
  2. Define Variables: Assign variables (like x and y) to the changing quantities.
  3. Write the Equation: Use the context to create a quadratic equation, often in standard form: f(x) = ax² + bx + c.
  4. Solve for the Question: Find the vertex for maximum/minimum values, or use factoring/the quadratic formula to find x-intercepts (roots).
  5. Interpret the Answer: Always state what your solution means in the original context of the problem.

Visual Examples

Example 1: Maximum Area

You have 60 feet of fencing for a rectangular garden against a wall. What dimensions maximize the area?

  1. Let the width be w. The length is 60 - 2w.
  2. Area A = length × width = w(60 - 2w).
  3. Equation: A = -2w² + 60w. (This is a parabola opening down).
  4. The maximum occurs at the vertex. w = -b/(2a) = -60/(2*-2) = 15 feet.
  5. Length = 60 - 2(15) = 30 feet. The maximum area is achieved with a width of 15 ft and length of 30 ft.

Example 2: Projectile Motion

A ball is thrown. Its height (in feet) is given by h(t) = -16t² + 48t + 5. When does it hit the ground?

  1. It hits the ground when height h(t) = 0.
  2. Set the equation to zero: 0 = -16t² + 48t + 5.
  3. Use the quadratic formula: t = [-48 ± √(48² - 4(-16)(5))] / (2*-16).
  4. This simplifies to t = [-48 ± √(2304 + 320)] / -32 = [-48 ± √2624] / -32.
  5. You get t ≈ -0.1 and t ≈ 3.1. Time cannot be negative, so the ball hits the ground at approximately 3.1 seconds.

Common Mistakes & How to Avoid Them

Ignoring the Context: A negative solution for time or length is not possible. Always check if your answers make sense in the real world.

Vertex Confusion: For a parabola opening down (a < 0), the vertex is the maximum point. For a parabola opening up (a > 0), it's the minimum. Check the sign of 'a'!

Misapplying the Formula: The vertex x-coordinate is -b/(2a). A common error is to forget the negative sign in front of the 'b'.

Tips & Tricks

Vertex Shortcut: The x-coordinate of the vertex is always x = -b/(2a). Plug this value back into the function to find the maximum or minimum y-value.

Memory Aid: The "a" in ax² tells you the parabola's direction. A positive "a" makes a happy smiley face (U-shape). A negative "a" makes a sad face (n-shape).

How to Practice

  • Start with simple area problems to build your equation-writing skills.
  • Practice identifying the vertex in different forms (standard, vertex).
  • Use graphing calculators or software to visualize the parabola and check your vertex and intercept answers.
  • Create your own word problems! This forces you to understand the components deeply.

Practice problems

6 of the 98, worked through step by step — try them before opening the answer.

1 A rectangular garden has a perimeter of 60 meters. If the area is maximized, what is the length of one side?

Hint: For a fixed perimeter, a rectangle's area is maximized when it becomes a square. Consider how to express area in terms of one variable using the perimeter constraint.

Show the answer

Answer: 15

  1. Let the length be L and width be W. The perimeter is 2L + 2W = 60.
  2. Simplify to L + W = 30, so W = 30 - L.
  3. The area A = L × W = L(30 - L) = 30L - L².
  4. This is a quadratic function A = -L² + 30L, which opens downward.
  5. The maximum occurs at the vertex L = -b/(2a) = -30/(2×-1) = 15.
  6. When L = 15, W = 30 - 15 = 15, so it's a square. The length of one side is 15 meters.

2 A company's profit P(x) = -2x² + 80x - 600, where x is units sold. Find the number of units that maximizes profit.

Hint: For a quadratic function in standard form, the vertex represents the maximum or minimum point. Consider the formula for finding the x-coordinate of the vertex.

Show the answer

Answer: 20

  1. The profit function is P(x) = -2x² + 80x - 600, which is a quadratic function in standard form ax² + bx + c.
  2. For a quadratic function, the x-coordinate of the vertex is given by x = -b/(2a).
  3. Here, a = -2 and b = 80.
  4. Substitute into the formula: x = -80/(2×(-2)) = -80/(-4) = 20.
  5. Since a = -2 is negative, the parabola opens downward, so the vertex represents a maximum point.
  6. Therefore, the number of units that maximizes profit is 20.

3 A rectangular garden has an area of 80 m². If the length is 4 meters more than the width, find the dimensions of the garden.

Hint: Set up a quadratic equation using the area formula for a rectangle and the relationship between length and width.

Show the answer

Answer: width = 8 m, length = 12 m

  1. Let width = w meters, then length = w + 4 meters
  2. Area = length × width = (w + 4) × w = 80
  3. Expand: w² + 4w = 80
  4. Rearrange: w² + 4w - 80 = 0
  5. Solve using quadratic formula: w = [-4 ± √(16 + 320)]/2 = [-4 ± √336]/2
  6. Simplify: w = [-4 ± 4√21]/2 = -2 ± 2√21
  7. Since width must be positive: w = -2 + 2√21 ≈ -2 + 9.165 = 7.165
  8. Check: 7.165 × 11.165 ≈ 80
  9. Exact solution: width = 2√21 - 2, length = 2√21 + 2
  10. For integer approximation: width ≈ 7.17 m, length ≈ 11.17 m The exact dimensions are width = 2√21 - 2 m, length = 2√21 + 2 m.

4 A company's profit P(x) = -2x² + 40x - 128, where x is units sold (in hundreds). Find the number of units that maximizes profit.

Hint: For a quadratic function in standard form, the vertex represents the maximum or minimum point. Consider how to find the x-coordinate of the vertex using the coefficients.

Show the answer

Answer: 1000

  1. The profit function is P(x) = -2x² + 40x - 128, which is a quadratic function in standard form ax² + bx + c.
  2. Since a = -2 is negative, the parabola opens downward, and the vertex represents the maximum point.
  3. The x-coordinate of the vertex is given by x = -b/(2a).
  4. Substitute b = 40 and a = -2: x = -40/(2×(-2)) = -40/(-4) = 10.
  5. Since x represents units sold in hundreds, the actual number of units is 10 × 100 = 1000. The number of units that maximizes profit is 1000.

5 A company's profit P(x) = -2x² + 80x - 600, where x is the number of units sold. Find the number of units that maximizes profit.

Hint: For a quadratic function in the form ax² + bx + c, the x-coordinate of the vertex gives the maximum or minimum value. Consider the formula that finds this x-coordinate.

Show the answer

Answer: 20

  1. The profit function is P(x) = -2x² + 80x - 600. Since the coefficient of x² is negative (-2), the parabola opens downward, and the vertex represents the maximum point.
  2. For a quadratic function in the form ax² + bx + c, the x-coordinate of the vertex is given by x = -b/(2a).
  3. Here, a = -2 and b = 80.
  4. Substitute into the formula: x = -80/(2 * -2) = -80/(-4) = 20.
  5. Therefore, the number of units that maximizes profit is 20.

6 A rectangular garden has a perimeter of 60 meters. If the area is modeled by A(w) = w(30 - w), find the maximum possible area of the garden.

Hint: For a quadratic function in the form ax² + bx + c, the maximum or minimum value occurs at the vertex. Consider how to find the vertex of a parabola.

Show the answer

Answer: 225

  1. The area function is A(w) = w(30 - w) = -w² + 30w
  2. This is a quadratic function opening downward (a = -1 < 0), so it has a maximum at the vertex
  3. The vertex occurs at w = -b/(2a) = -30/(2×-1) = -30/-2 = 15
  4. Substitute w = 15 into the area function: A(15) = 15(30 - 15) = 15 × 15 = 225
  5. The maximum possible area is 225 square meters.
Practise this topic — 10 free problems, no signup →