WealthWiseGuys Practical Mathematics

Chapter 25

Functions, Quadratic Equations, and Maximum Profit

Connect inputs to outputs, solve curves, and evaluate a modeled maximum

A function organizes a relationship so that each allowed input has one output. Quadratic functions add curves, allowing you to model changing revenue and locate a maximum under stated assumptions. All businesses, prices, and earnings are hypothetical; money is rounded to the nearest cent when needed, and modeled profit is not guaranteed.

25.1

Functions, Notation, Domain, and Range

Learning objectives

After this section you will be able to:

  1. Determine whether a relationship is a function.
  2. Evaluate functions from formulas and tables.
  3. Identify meaningful domains and corresponding ranges.

Real-life situation

A creator workshop fee

A hypothetical workshop charges a $25 booking fee plus $12 per participant. If n participants attend, f(n) gives the total charge. Each allowed participant count produces one charge.

Participants nCharge f(n) ($)
025
249
585
Figure 25.1 Values of the hypothetical function f(n) = 25 + 12n

The concept

A function assigns exactly one output to each allowed input. Several inputs may have the same output, but one input cannot have two different outputs in the same function. A table with the same input paired with conflicting outputs does not describe a function.

Function notation names the rule and its input. Read f(4) as 'f of four,' not f multiplied by four. To evaluate it, replace every occurrence of the input variable with 4 and follow the order of operations.

The domain is the collection of permitted inputs; the range is the collection of outputs actually produced. A formula alone does not settle a real-world domain. A workshop with room for 20 participants uses n = 0, 1, 2, ..., 20, not negative values or fractional people.

A graph represents a function when no vertical line intersects it more than once. Each horizontal location then has at most one output. Tables and formulas also help distinguish linear functions, which have constant first differences for equally spaced inputs, from nonlinear functions, whose changes may vary.

Functions, Notation, Domain, and Range: solution map

  1. 1

    Identify the question

    For f(n) = 25 + 12n, find f(4), then find the participant count producing a charge of $85.

  2. 2

    Calculation 1

    Substitute 4: f(4) = 25 + 12 × 4.

  3. 3

    Calculation 2

    Calculate: f(4) = 25 + 48 = 73.

  4. 4

    Calculation 3

    For an $85 charge, write 25 + 12n = 85.

  5. 5

    Calculation 4

    Subtract 25 and divide by 12: n = 60 ÷ 12 = 5.

  6. 6

    Calculation 5

    Check: f(5) = 25 + 60 = 85.

  7. 7

    Interpret and check

    Four participants cost $73.00; an $85.00 charge corresponds to 5 participants.

Figure 25.1.V Functions, Notation, Domain, and Range: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • profit

    What is left when expenses are subtracted from revenue.

  • profit margin

    Profit shown as a percent of revenue; a 10% margin means $10 of profit per $100 of sales.

  • revenue

    The total money a business brings in from sales, before any costs are paid.

Key terms

Function
A rule assigning exactly one output to each allowed input.
Function notation
Notation such as f(x), naming a function and its input.
Domain
The allowed input values.
Range
The outputs produced by the allowed inputs.
Vertical-line test
A graph is a function if every vertical line meets it at most once.

Formulas & rules

  • Workshop charge

    f(n) = 25 + 12n

  • Evaluation pattern

    If f(x) = ax + b, then f(k) = a × k + b

  • A nonlinear rule

    g(x) = 2x² + 5

    Square the input before multiplying by 2.

Worked examples

Example 1Evaluate and reverse a function

For f(n) = 25 + 12n, find f(4), then find the participant count producing a charge of $85.

Solution

  1. 1Substitute 4: f(4) = 25 + 12 × 4.
  2. 2Calculate: f(4) = 25 + 48 = 73.
  3. 3For an $85 charge, write 25 + 12n = 85.
  4. 4Subtract 25 and divide by 12: n = 60 ÷ 12 = 5.
  5. 5Check: f(5) = 25 + 60 = 85.

Answer: Four participants cost $73.00; an $85.00 charge corresponds to 5 participants.

Example 2Evaluate a nonlinear function

Find g(3) for g(x) = 2x² + 5.

Solution

  1. 1Replace x with 3: g(3) = 2 × 3² + 5.
  2. 2Square first: 3² = 9.
  3. 3Multiply: 2 × 9 = 18.
  4. 4Add: 18 + 5 = 23.

Answer: g(3) = 23.

Practice exercises 25.1

  1. Find f(0) and f(3) for f(n) = 25 + 12n.
  2. Solve f(n) = 61 for the workshop function.
  3. Find g(4) for g(x) = 2x² + 5.
  4. Does {(1, 8), (2, 8), (3, 10)} represent a function?
  5. For g(x) = 2x² + 5 with domain {0, 2}, find the range.

Recap quiz 25.1

1. For f(n) = 25 + 12n, f(2) is:
2. For h(x) = 3x + 1, h(2) is:
3. Which table fails to represent a function?

Section 25.1 recap

  • A function gives one output per allowed input; f(x) is notation, not multiplication.
  • Use the situation to restrict the domain, then calculate the resulting range.
Answer key with explanations: Section 25.1

Practice exercises

  1. f(0) = 25 + 0 = $25.00; f(3) = 25 + 36 = $61.00. The zero-input charge is the booking fee.
  2. 25 + 12n = 61 gives 12n = 36, so n = 3 participants.
  3. g(4) = 2 × 16 + 5 = 32 + 5 = 37.
  4. Yes. Each input appears with exactly one output. Inputs 1 and 2 may share the output 8.
  5. g(0) = 5 and g(2) = 2 × 4 + 5 = 13, so the range is {5, 13}.

Recap quiz

  1. $49: 25 + 12 × 2 = 25 + 24 = $49.00.
  2. 7: 3 × 2 + 1 = 7.
  3. (2, 3), (2, 7): Input 2 is assigned two different outputs, 3 and 7.
25.2

Quadratic Equations and Break-Even Roots

Learning objectives

After this section you will be able to:

  1. Recognize quadratic expressions and equations.
  2. Solve quadratics by factoring or the quadratic formula.
  3. Interpret roots and reject solutions outside the domain.

Real-life situation

A curved profit model

A hypothetical small batch has modeled profit P(q) = −q² + 10q − 16 dollars. Unlike a constant-margin model, profit eventually falls as quantity rises. The zero-profit quantities separate modeled losses from gains.

Quantity qProfit P(q) ($)
0-16
20
59
80
Figure 25.2 Values of a hypothetical quadratic profit model

The concept

A quadratic has a nonzero squared term and can be written ax² + bx + c. Its graph is a parabola. Positive a produces an upward-opening curve, while negative a produces a downward-opening curve.

A root is an input making the function zero. To solve a quadratic equation, first move all terms to one side. When an expression factors, the zero-product property says at least one factor must equal zero.

Not every quadratic factors conveniently. The quadratic formula works for any quadratic equation with a ≠ 0. Substitute signed coefficients carefully, especially when b or c is negative, and keep exact values until the final rounding step.

The discriminant b² − 4ac tells how many real roots exist: positive gives two distinct roots, zero gives one repeated root, and negative gives no real roots. A model with no real zero need not be meaningless; it simply does not cross the horizontal axis.

Interpret each root in context. A negative quantity may be algebraically valid but unusable. For a downward-opening profit model with two valid roots, profit is positive between them and negative outside them, within the model's permitted domain.

Quadratic Equations and Break-Even Roots: solution map

  1. 1

    Identify the question

    Find the zero-profit quantities for P(q) = −q² + 10q − 16.

  2. 2

    Calculation 1

    Set profit equal to zero: −q² + 10q − 16 = 0.

  3. 3

    Calculation 2

    Multiply by −1: q² − 10q + 16 = 0.

  4. 4

    Calculation 3

    Find numbers multiplying to 16 and adding to −10: −2 and −8.

  5. 5

    Calculation 4

    Factor: (q − 2)(q − 8) = 0.

  6. 6

    Calculation 5

    Set each factor to zero: q = 2 or q = 8. Substitution gives zero profit at both.

  7. 7

    Interpret and check

    The model breaks even at 2 and 8 units; its profit is positive for 2 < q < 8.

Figure 25.2.V Quadratic Equations and Break-Even Roots: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • fixed cost

    A cost that stays the same no matter how many items are sold, like rent.

  • variable cost

    A cost that rises or falls with the number of items made or sold.

  • net profit

    The final profit after every cost is paid — sometimes called the bottom line.

Key terms

Quadratic
An expression or function whose highest variable power is 2.
Parabola
The curved graph of a quadratic function.
Root
An input at which a function equals zero.
Factor
An expression multiplied by another to form a product.
Discriminant
The value b² − 4ac used to classify a quadratic's real roots.

Formulas & rules

  • Standard quadratic equation

    ax² + bx + c = 0

    a ≠ 0.

  • Quadratic formula

    x = (−b ± √(b² − 4ac)) ÷ (2a)

  • Zero-product property

    If A × B = 0, then A = 0 or B = 0

  • Discriminant

    D = b² − 4ac

Worked examples

Example 1Factor to find break-even

Find the zero-profit quantities for P(q) = −q² + 10q − 16.

Solution

  1. 1Set profit equal to zero: −q² + 10q − 16 = 0.
  2. 2Multiply by −1: q² − 10q + 16 = 0.
  3. 3Find numbers multiplying to 16 and adding to −10: −2 and −8.
  4. 4Factor: (q − 2)(q − 8) = 0.
  5. 5Set each factor to zero: q = 2 or q = 8. Substitution gives zero profit at both.

Answer: The model breaks even at 2 and 8 units; its profit is positive for 2 < q < 8.

Example 2Use the quadratic formula

Solve q² − 10q + 16 = 0 using the quadratic formula.

Solution

  1. 1Identify a = 1, b = −10, and c = 16.
  2. 2Compute D = (−10)² − 4 × 1 × 16 = 100 − 64 = 36.
  3. 3Substitute: q = (10 ± √36) ÷ 2 = (10 ± 6) ÷ 2.
  4. 4The smaller root is (10 − 6) ÷ 2 = 2.
  5. 5The larger root is (10 + 6) ÷ 2 = 8.

Answer: q = 2 or q = 8, matching the factored solution.

Practice exercises 25.2

  1. Solve x² − 7x + 12 = 0.
  2. Solve x² − 16 = 0. If x is a length, which root is usable?
  3. Solve 2x² − 10x + 12 = 0.
  4. How many real roots does x² + 4x + 5 = 0 have?
  5. Verify that x = 3 solves x² − 5x + 6 = 0.

Recap quiz 25.2

1. The solutions of x² − 9 = 0 are:
2. A positive discriminant means:
3. Which equation is quadratic?

Section 25.2 recap

  • Set the expression equal to zero before using factors or the quadratic formula.
  • Check roots algebraically and against the situation's domain.
Answer key with explanations: Section 25.2

Practice exercises

  1. Factor as (x − 3)(x − 4) = 0 because −3 + (−4) = −7 and (−3)(−4) = 12. Thus x = 3 or 4.
  2. (x − 4)(x + 4) = 0 gives x = 4 or −4. A length must be nonnegative, so only 4 is usable.
  3. Divide by 2: x² − 5x + 6 = 0. Factor (x − 2)(x − 3) = 0, giving x = 2 or 3.
  4. D = 4² − 4 × 1 × 5 = 16 − 20 = −4. The negative discriminant means no real roots.
  5. Substitute 3: 3² − 5 × 3 + 6 = 9 − 15 + 6 = 0, so it is a root.

Recap quiz

  1. −3 and 3: (x − 3)(x + 3) = 0 gives both roots.
  2. Two distinct real roots: A positive square root produces two different values in the quadratic formula.
  3. x² − 5x + 6 = 0: Its highest variable power is 2 and its squared-term coefficient is nonzero.
25.3

The Vertex and Maximum Modeled Profit

Learning objectives

After this section you will be able to:

  1. Build a quadratic profit function from price and cost assumptions.
  2. Find and evaluate a parabola's vertex.
  3. Apply capacity and whole-number restrictions to an optimization result.

Real-life situation

Pricing a hypothetical community event

A hypothetical Philadelphia student event models ticket price as p(q) = 30 − 0.20q dollars when q tickets sell. Its cost is $200 plus $10 per attendee. The assumed price-demand relationship creates a quadratic profit model, not a guaranteed sales forecast.

Tickets qModeled profit ($)
0-200
10-20
30220
50300
70220
Figure 25.3 Hypothetical event profit P(q) = −0.20q² + 20q − 200

The concept

When the selling price depends on quantity, revenue need not be linear. Multiplying p(q) = 30 − 0.20q by q gives R(q) = 30q − 0.20q². Subtracting C(q) = 200 + 10q produces P(q) = −0.20q² + 20q − 200.

A parabola's vertex is its turning point. For f(x) = ax² + bx + c, the vertex input is −b ÷ (2a). When a is negative, the parabola opens downward and the vertex gives its unrestricted maximum. When a is positive, the vertex is a minimum.

Calculate the output by substituting the vertex input into the original function. In a business problem, also recover the corresponding price, revenue, and cost so the final recommendation is understandable and can be checked.

Optimization must respect the domain. Here nonnegative ticket price requires q ≤ 150, and actual attendance uses whole numbers. Capacity may impose a smaller upper bound. For a continuous quadratic on a closed interval, compare an allowed vertex with the interval endpoints; for whole-number quantities, compare the allowed integers closest to the vertex and any relevant boundaries.

The result is only as dependable as the assumptions. Actual demand, refunds, staffing costs, and capacity may differ. Maximum modeled profit is neither investment advice nor a promise that a business or event will earn that amount.

The Vertex and Maximum Modeled Profit: solution map

  1. 1

    Identify the question

    For the event, find the quantity, price, and profit at the unrestricted vertex. Assume capacity is at least 50.

  2. 2

    Calculation 1

    Build profit: P(q) = q(30 − 0.20q) − (200 + 10q) = −0.20q² + 20q − 200.

  3. 3

    Calculation 2

    With a = −0.20 and b = 20, qᵥ = −20 ÷ (2 × −0.20) = 50.

  4. 4

    Calculation 3

    Price: p(50) = 30 − 0.20 × 50 = $20.00.

  5. 5

    Calculation 4

    Revenue: R = 50 × 20 = $1,000.00.

  6. 6

    Calculation 5

    Cost: C = 200 + 10 × 50 = $700.00; profit = 1,000 − 700 = $300.00.

  7. 7

    Calculation 6

    Since a < 0 and 50 is an allowed whole-number quantity, the vertex gives the modeled maximum.

  8. 8

    Interpret and check

    The model's maximum is $300.00 at 50 tickets priced at $20.00 each.

Figure 25.3.V The Vertex and Maximum Modeled Profit: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • cost

    The amount paid to make or buy an item before it is sold.

  • selling price

    The price a customer pays for an item or service.

  • break-even point

    The sales level where revenue exactly equals total cost, so there is no profit and no loss.

Key terms

Vertex
The turning point of a parabola.
Axis of symmetry
The vertical line through the vertex, dividing a parabola into mirror halves.
Maximum
The greatest output over the allowed domain.
Minimum
The smallest output over the allowed domain.
Price-demand model
An assumed relationship between selling price and quantity sold.

Formulas & rules

  • Revenue with changing price

    R(q) = q × p(q)

  • Profit

    P(q) = R(q) − C(q)

  • Vertex input

    xᵥ = −b ÷ (2a)

    For ax² + bx + c, a ≠ 0.

  • Vertex output

    yᵥ = a × xᵥ² + b × xᵥ + c

Worked examples

Example 1Find the modeled profit maximum

For the event, find the quantity, price, and profit at the unrestricted vertex. Assume capacity is at least 50.

Solution

  1. 1Build profit: P(q) = q(30 − 0.20q) − (200 + 10q) = −0.20q² + 20q − 200.
  2. 2With a = −0.20 and b = 20, qᵥ = −20 ÷ (2 × −0.20) = 50.
  3. 3Price: p(50) = 30 − 0.20 × 50 = $20.00.
  4. 4Revenue: R = 50 × 20 = $1,000.00.
  5. 5Cost: C = 200 + 10 × 50 = $700.00; profit = 1,000 − 700 = $300.00.
  6. 6Since a < 0 and 50 is an allowed whole-number quantity, the vertex gives the modeled maximum.

Answer: The model's maximum is $300.00 at 50 tickets priced at $20.00 each.

Example 2Check symmetry and a capacity limit

Compare profit at 40 and 60 tickets. If capacity is only 40, what quantity maximizes modeled profit?

Solution

  1. 1At 40: P(40) = −0.20 × 1,600 + 800 − 200 = $280.00.
  2. 2At 60: P(60) = −0.20 × 3,600 + 1,200 − 200 = $280.00.
  3. 3These quantities are equally far from the vertex q = 50, so their outputs match.
  4. 4With capacity 40, the allowed domain is whole-number q from 0 through 40.
  5. 5The quadratic increases up to q = 50, so its greatest allowed value occurs at q = 40.

Answer: Both quantities produce $280.00 in the model; with capacity 40, choose 40 tickets for the maximum feasible modeled profit.

Practice exercises 25.3

  1. Find the vertex and maximum of P(q) = −q² + 12q − 20 for q ≥ 0.
  2. Find the maximum of P(q) = −2q² + 24q − 40 for q ≥ 0.
  3. Does f(x) = x² + 8x + 3 have an unrestricted maximum or minimum?
  4. For the event model, calculate P(41). Why is it unusable if capacity is 40?
  5. For P(q) = −q² + 7q and whole-number q ≥ 0, find the maximizing quantities.

Recap quiz 25.3

1. For P(q) = −q² + 10q, the vertex quantity is:
2. What is the maximum value of P(q) = −q² + 10q for q ≥ 0?
3. A vertex quantity exceeds capacity. What should you do?

Section 25.3 recap

  • For a downward-opening quadratic, an allowed vertex gives the maximum.
  • Check price, capacity, whole-number quantities, and modeling assumptions before interpreting the result.
Answer key with explanations: Section 25.3

Practice exercises

  1. qᵥ = −12 ÷ (−2) = 6. P(6) = −36 + 72 − 20 = 16. Since a < 0, the maximum is 16 at q = 6.
  2. qᵥ = −24 ÷ (−4) = 6. P(6) = −72 + 144 − 40 = 32, the maximum.
  3. Since a = 1 > 0, it has a minimum. The vertex input is −8 ÷ 2 = −4; f(−4) = 16 − 32 + 3 = −13. There is no unrestricted maximum.
  4. P(41) = −0.20 × 1,681 + 820 − 200 = −336.20 + 620 = $283.80. It exceeds P(40), but 41 attendees violate the capacity constraint.
  5. The continuous vertex is q = −7 ÷ (−2) = 3.5. Check adjacent integers: P(3) = −9 + 21 = 12 and P(4) = −16 + 28 = 12. Both 3 and 4 maximize the whole-number model.

Recap quiz

  1. 5: qᵥ = −10 ÷ (2 × −1) = 5.
  2. 25: At q = 5, P = −25 + 50 = 25; the downward-opening parabola has its maximum there.
  3. Optimize over the allowed domain: An infeasible vertex cannot be chosen; evaluate the objective within the actual constraints.

Chapter 25 Organizer and Review

Chapter Organizer

Every section at a glance: key formulas, key terms, and the main idea.

  • 25.1 Functions, Notation, Domain, and Range

    Key formulas

    • Workshop charge: f(n) = 25 + 12n
    • Evaluation pattern: If f(x) = ax + b, then f(k) = a × k + b
    • A nonlinear rule: g(x) = 2x² + 5

    Key terms

    Function, Function notation, Domain, Range, Vertical-line test

    Remember

    A function gives one output per allowed input; f(x) is notation, not multiplication.

  • 25.2 Quadratic Equations and Break-Even Roots

    Key formulas

    • Standard quadratic equation: ax² + bx + c = 0
    • Quadratic formula: x = (−b ± √(b² − 4ac)) ÷ (2a)
    • Zero-product property: If A × B = 0, then A = 0 or B = 0
    • Discriminant: D = b² − 4ac

    Key terms

    Quadratic, Parabola, Root, Factor, Discriminant

    Remember

    Set the expression equal to zero before using factors or the quadratic formula.

  • 25.3 The Vertex and Maximum Modeled Profit

    Key formulas

    • Revenue with changing price: R(q) = q × p(q)
    • Profit: P(q) = R(q) − C(q)
    • Vertex input: xᵥ = −b ÷ (2a)
    • Vertex output: yᵥ = a × xᵥ² + b × xᵥ + c

    Key terms

    Vertex, Axis of symmetry, Maximum, Minimum, Price-demand model

    Remember

    For a downward-opening quadratic, an allowed vertex gives the maximum.

Cumulative Review

Mixed practice from this chapter and the two chapters before it.

  1. Section 25.1

    For g(x) = 2x² + 5 with domain {0, 2}, find the range.

    Show answer

    g(0) = 5 and g(2) = 2 × 4 + 5 = 13, so the range is {5, 13}.

  2. Section 25.2

    Verify that x = 3 solves x² − 5x + 6 = 0.

    Show answer

    Substitute 3: 3² − 5 × 3 + 6 = 9 − 15 + 6 = 0, so it is a root.

  3. Section 25.3

    For P(q) = −q² + 7q and whole-number q ≥ 0, find the maximizing quantities.

    Show answer

    The continuous vertex is q = −7 ÷ (−2) = 3.5. Check adjacent integers: P(3) = −9 + 21 = 12 and P(4) = −16 + 28 = 12. Both 3 and 4 maximize the whole-number model.

  4. Review of Section 23.3

    Find the stand's profit at 40 items.

    Show answer

    P = (12 − 8) × 40 − 80 = 160 − 80 = $80.00.

  5. Review of Section 24.3

    Is (3, 5) feasible for the shop?

    Show answer

    Total packs: 3 + 5 = 8 ≤ 8. Machine time: 2 × 3 + 5 = 11 ≤ 12. Both counts are nonnegative integers, so yes.

All prices, stores, people, and accounts in this chapter are hypothetical and for teaching only. Product references are generic categories; no brand endorses or is endorsed by this textbook.