Quadratic and Exponential Word Problems: Foundations

[Video: Solving Quadratic and Exponential Word Problems]

Learn how to translate real-world scenarios into quadratic and exponential equations and solve them systematically.

Problem-Solving Approach

General Strategy

  1. Understand the problem scenario
  2. Identify what's given and what's asked
  3. Choose the appropriate model (quadratic or exponential)
  4. Translate words into mathematical equations
  5. Solve the equation(s)
  6. Interpret the solution in context

Quadratic Word Problems

Common Scenarios

Example: A ball is thrown upward from a height of 5 feet with an initial velocity of 48 ft/s. Its height h after t seconds is given by h = -16t² + 48t + 5. What is the maximum height reached?

Solution: The maximum occurs at the vertex. t = -b/(2a) = -48/(2×-16) = 1.5 seconds. Then h = -16(1.5)² + 48(1.5) + 5 = 41 feet.

Exponential Word Problems

Common Scenarios

Example: A bacteria culture starts with 500 bacteria and doubles every hour. How many bacteria will there be after 6 hours?

Solution: Exponential growth model: N = 500 × 2^t. After 6 hours: N = 500 × 2^6 = 500 × 64 = 32,000 bacteria.

Identifying Model Types

Keywords Model Type
"maximum," "minimum," "vertex," "parabola" Quadratic
"doubles," "triples," "half-life," "grows by percentage" Exponential
"constant rate" or "linear change" Linear

Key Takeaways

Practice Question

A car's value depreciates exponentially over time. The car was originally purchased for $25,000 and was worth $20,000 after 1 year. What will its value be after 3 years? (Round to the nearest dollar.)

A) $12,800
B) $13,107
C) $15,000
D) $16,000

Follow these steps:

  1. Determine the annual decay factor from the given values
  2. Create the exponential decay model V = V₀ × (decay factor)^t
  3. Use the model to calculate the value after 3 years
  4. Round to the nearest dollar

Step-by-Step Solution:

1. Find decay factor: 20,000 = 25,000 × r^1 → r = 0.8

2. Model: V = 25,000 × (0.8)^t

3. After 3 years: V = 25,000 × (0.8)^3 = 25,000 × 0.512 = $12,800

The correct answer is A) $12,800.