8.10Quadratic Applications and Modeling
Apply quadratic functions to real-world situations — projectile motion, maximum area, revenue, and optimization — by interpreting the vertex, zeros, y-intercept, and domain in context.
Why This Matters
Quadratic models appear on the SAT and ACT in word problems involving projectile motion, area, and revenue. Understanding how to extract meaning from a quadratic function — not just solve it — is essential for success in Algebra 2, Physics, and Economics.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Quick Answer
How Do You Use a Quadratic Function to Model a Real-World Situation?
Define the variables, write or identify a quadratic function, and determine which feature answers the question. Use the vertex for a maximum or minimum, the zeros for when the output equals zero, and selected function values for other conditions. Always interpret the result using the situation, domain, and units.
Learning Goals
- Identify when a real-world situation can be modeled by a quadratic function.
- Write a quadratic model from a word problem (area, projectile, revenue).
- Use the vertex to find a maximum or minimum value in context.
- Use the zeros to find when a quantity equals zero (e.g., landing time).
- Interpret the y-intercept as an initial value.
- Restrict the domain to values that are physically meaningful.
- Reject extraneous solutions that do not fit the context.
Key Vocabulary
Quadratic Model
A quadratic function used to represent a real-world situation, such as height over time or area as a function of a dimension.
Example: h(t) = −16t² + 64t
Vertex (in context)
The point (h, k) where the parabola reaches its maximum (a < 0) or minimum (a > 0). In context, k is the maximum height, minimum cost, etc.
Example: Vertex (2, 64): maximum height 64 ft at t = 2 s
Zeros / x-intercepts (in context)
The values of x where f(x) = 0. In projectile problems, these are the times when the object is at height zero (launch and landing).
Example: t = 0 and t = 4 in h(t) = −16t² + 64t
y-intercept (in context)
The value of f(0) = c. In a height model, this is the initial height at time zero.
Example: h(0) = 48 means the object starts 48 ft above the ground
Maximum / Minimum
The largest (maximum) or smallest (minimum) output value of a quadratic function. Found at the vertex.
Domain (reasonable domain)
The set of input values that make sense in context. For time, the domain is t ≥ 0. For length, x > 0 and x must not exceed physical limits.
Projectile Motion
The path of an object launched into the air under gravity, modeled by h(t) = −16t² + v₀t + h₀ (feet) or h(t) = −4.9t² + v₀t + h₀ (meters).
Revenue / Profit
Revenue = price × quantity. Profit = Revenue − Cost. Both can be modeled by quadratic functions when price or quantity is a variable.
Recognizing Quadratic Patterns
A situation calls for a quadratic model when the relationship involves a product of two linear expressions (e.g., length × width where one dimension depends on the other), a square (e.g., area of a square side), or a constant second difference in a table. Common contexts include:
Projectile motion
Height is a quadratic function of time: h(t) = −16t² + v₀t + h₀
Area optimization
Area = length × width, where one dimension is expressed in terms of the other
Revenue / Profit
Revenue = price × quantity; when quantity depends on price, the product is quadratic
Important Features in Context
Context Question → Feature to Use
Maximum height / minimum cost
Vertex (h, k)
When does it hit the ground?
Zeros (x-intercepts)
Starting value / initial height
y-intercept (0, c)
Height at t = 3 seconds
Evaluate f(3)
The Modeling Process
The 7-Step Modeling Process
- 1Define variables — what does x represent? What does f(x) represent?
- 2Identify given information (numbers, constraints, units).
- 3Write or select a quadratic model.
- 4Choose the required feature or solving method (vertex, zeros, evaluate).
- 5Perform the calculation.
- 6Interpret the answer in context.
- 7Check domain, units, and reasonableness.
Three Forms — When to Use Each
Standard form
y = ax² + bx + c
Best for: y-intercept, solving by formula or factoring
Vertex form
y = a(x − h)² + k
Best for: Maximum or minimum value, graphing
Factored form
y = a(x − r₁)(x − r₂)
Best for: Zeros (x-intercepts)
Worked Examples
A ball is launched from the ground: h(t) = −16t² + 64t. (a) Find the maximum height. (b) When does it hit the ground?
(a) Maximum height — use the vertex.
Axis of symmetry: t = −b/(2a) = −64/(2·−16) = −64/−32 = 2
h(2) = −16(2)² + 64(2) = −16(4) + 128 = −64 + 128 = 64
Maximum height: 64 feet at t = 2 seconds.
(b) Hits the ground — find the zeros.
Set h(t) = 0: −16t² + 64t = 0
Factor: −16t(t − 4) = 0
t = 0 (launch) or t = 4
Reject t = 0 (that is the launch point). The ball hits the ground at t = 4 seconds.
A ball is thrown from a 48-ft platform: h(t) = −16t² + 32t + 48. Find the maximum height and when it hits the ground.
Maximum height — vertex.
t = −32/(2·−16) = −32/−32 = 1
h(1) = −16(1) + 32(1) + 48 = −16 + 32 + 48 = 64
Maximum height: 64 feet at t = 1 second.
Hits the ground — set h(t) = 0.
−16t² + 32t + 48 = 0
Divide by −16: t² − 2t − 3 = 0
Factor: (t − 3)(t + 1) = 0
t = 3 or t = −1
Reject t = −1 (time cannot be negative). The ball hits the ground at t = 3 seconds.
In Example 2, why is t = −1 rejected?
The variable t represents time elapsed since the ball was thrown.
Time cannot be negative — the ball does not exist before it is thrown.
The reasonable domain is t ≥ 0.
Therefore t = −1 is an extraneous solution and is rejected.
Only t = 3 is meaningful in this context.
A farmer has 120 ft of fencing for 3 sides of a rectangular pen (one side is a barn wall). Find the dimensions that maximize area.
Let x = width of the pen (the two sides perpendicular to the barn).
The length (parallel to the barn) = 120 − 2x.
Area: A(x) = x(120 − 2x) = −2x² + 120x
Vertex: x = −120/(2·−2) = −120/−4 = 30
Length = 120 − 2(30) = 60
Maximum area = 30 × 60 = 1800 sq ft.
Check domain: x > 0 and 120 − 2x > 0 → x < 60. So 0 < x < 60. x = 30 is valid.
Rectangle Area Model
One side is a barn wall; fencing covers the other three sides: 2x + length = 120.
A rectangle has area 54 sq ft. Its length is (x + 3) and width is (x − 3). Find x and the dimensions.
Set up the equation: (x + 3)(x − 3) = 54
Difference of squares: x² − 9 = 54
x² = 63
x = √63 = 3√7 ≈ 7.94
Length = 7.94 + 3 ≈ 10.94 ft
Width = 7.94 − 3 ≈ 4.94 ft
Check: 10.94 × 4.94 ≈ 54 sq ft. ✓
Reject x = −3√7 because width x − 3 would be negative.
A 10-ft by 12-ft garden has a uniform border of width x. The planted area is 80 sq ft. Find x.
The planted region has dimensions (10 − 2x) by (12 − 2x).
Set up: (10 − 2x)(12 − 2x) = 80
Expand: 120 − 20x − 24x + 4x² = 80
4x² − 44x + 120 = 80
4x² − 44x + 40 = 0
Divide by 4: x² − 11x + 10 = 0
Factor: (x − 10)(x − 1) = 0
x = 10 or x = 1
Reject x = 10 (the border would exceed the garden width of 10 ft).
The border width is x = 1 ft.
Revenue is R(x) = −5x² + 200x where x is the price in dollars. Find the price that maximizes revenue and the maximum revenue.
Vertex: x = −200/(2·−5) = −200/−10 = 20
R(20) = −5(20)² + 200(20) = −5(400) + 4000 = −2000 + 4000 = 2000
Maximum revenue is $2000 at a price of $20.
Profit is P(x) = −2x² + 80x − 600 where x is units sold. Find the break-even points and maximum profit.
Break-even: set P(x) = 0.
−2x² + 80x − 600 = 0
Divide by −2: x² − 40x + 300 = 0
Factor: (x − 10)(x − 30) = 0
x = 10 or x = 30 (break-even points)
Maximum profit — vertex: x = −80/(2·−2) = −80/−4 = 20
P(20) = −2(400) + 80(20) − 600 = −800 + 1600 − 600 = 200
Maximum profit is $200 at x = 20 units.
Two consecutive positive integers have a product of 72. Find them.
Let n = the smaller integer. Then n + 1 = the larger.
n(n + 1) = 72
n² + n − 72 = 0
Factor: (n + 9)(n − 8) = 0
n = −9 or n = 8
Reject n = −9 (both integers must be positive).
n = 8, so the integers are 8 and 9.
Check: 8 × 9 = 72. ✓
Common Mistakes
Reporting both roots when only one fits context: t=4 or t=−1 → both are answers.
Reject t=−1 because time cannot be negative. Only t=4 is meaningful.
Ignoring units: 'The maximum is 64'.
Always state units: 'The maximum height is 64 feet.'
Confusing x- and y-values of the vertex: 'The maximum occurs at height 2'.
The vertex is (2, 64): the maximum height is 64 ft, occurring at t = 2 seconds.
Using the y-intercept when the problem asks for a maximum.
The y-intercept gives the starting value (t=0). The maximum is the vertex y-value.
Treating all quadratic contexts as projectile problems.
Quadratic models also appear in area, revenue, and optimization — read the context carefully.
Forgetting to restrict the domain: reporting x=−9 for a length or time.
Check that all answers are physically meaningful. Reject negative lengths, times, or quantities.
Rounding too early in multi-step problems.
Keep exact values through intermediate steps; round only in the final answer.
Writing a model without defining variables.
Always state what x and f(x) represent, including units, before writing the equation.
Assuming every table is quadratic without checking second differences.
Verify second differences are constant before fitting a quadratic model.
Reporting an impossible answer without checking reasonableness.
Ask: does this answer make sense in context? A 500-foot jump or negative profit should raise a flag.
Guided Practice
Guided Practice Video: Quadratic Applications and Modeling
Review setting up and solving real-world quadratic models — projectile motion, area problems, and number problems — before completing the guided problems below.
Video by Sang Real Math
Watch on YouTube ↗A ball is thrown upward: h(t) = −16t² + 48t. Find the maximum height and when it hits the ground.
Hint: Vertex: t = −48/(2·−16) = 1.5. Then h(1.5) = ? For the ground, set h(t) = 0 and factor.
A rectangle has perimeter 40 ft. Write the area as a function of width x, then find the dimensions that maximize area.
Hint: Perimeter: 2x + 2L = 40 → L = 20 − x. Area: A(x) = x(20 − x). Find the vertex.
Revenue is R(x) = −3x² + 120x. Find the price that maximizes revenue.
Hint: Vertex: x = −120/(2·−3). What is the maximum revenue?
Two consecutive positive integers have a product of 56. Find them.
Hint: Let n and n+1 be the integers. Set n(n+1) = 56. Expand, rearrange, factor. Reject any negative solution.
Related Lessons
Practice Questions
Interactive Practice — 5 Questions
A ball is thrown: h(t) = −16t² + 64t. The maximum height is:
For h(t) = −16t² + 64t, the ball hits the ground at t =
Revenue R(x) = −x² + 60x is maximized at x =
The y-intercept of a quadratic height model represents:
Which feature of a quadratic model gives the maximum or minimum value?