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Lesson 4 — Quadratic function

Quadratic function f(x) = ax² + bx + c. Discriminant, roots (quadratic formula), vertex form, vertex, concavity, axis of symmetry, sign analysis, maximum and minimum. Optimization of area, cost, and profit.

Used in: 1.º ano do EM (15–16 anos) · Math I japonês cap. 2 · Klasse 10 alemã · O-level Singapore cap. 2 · ENEM

f(x) = ax^2 + bx + c, \quad a \neq 0

Quadratic function — polynomial of degree 2. The graph is a parabola: opens upward if $a > 0$ , downward if $a < 0$ . The highest or lowest point is the vertex $V = (-b/(2a),\ -\Delta/(4a))$ .

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition

"A quadratic function is a polynomial function of degree 2. The graph of a quadratic function is a parabola." — OpenStax College Algebra 2e, §5.1

Roots — Quadratic formula

The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots of $ax^2 + bx + c = 0$ :

x = \frac{-b \pm \sqrt{\Delta}}{2a}, \quad \Delta = b^2 - 4ac

what this means · Quadratic formula: Δ > 0 → two distinct real roots; Δ = 0 → one double real root; Δ < 0 → no real roots (two complex conjugates).

Vertex and vertex form

Completing the square in $ax^2 + bx + c$ , we obtain the vertex form:

f(x) = a(x - x_V)^2 + y_V

what this means · Vertex form: directly reveals vertex V = (xV, yV), axis of symmetry x = xV, and extreme value yV.

where the vertex coordinates are:

x_V = -\frac{b}{2a}, \qquad y_V = f(x_V) = -\frac{\Delta}{4a}

what this means · xV is also the average of the roots (when they exist). yV is the minimum value if a > 0, maximum if a < 0.

Parabolas — geometric figure

Left: a > 0, opens upward, vertex is minimum. Right: a < 0, opens downward, vertex is maximum. Orange points: roots (zeros of the function). Dashed line: axis of symmetry.

Vieta's formulas

If $x_1, x_2$ are the roots of $ax^2 + bx + c = 0$ (when $\Delta \geq 0$ ):

x_1 + x_2 = -\frac{b}{a}, \qquad x_1 \cdot x_2 = \frac{c}{a}

what this means · Girard-Vieta relations: sum and product of roots expressed directly in terms of coefficients, without needing to find the roots individually.

"The vertex of the parabola is the maximum point if $a < 0$ or the minimum point if $a > 0$ ." — OpenStax College Algebra 2e, §5.1

Sign of the quadratic function

With roots $x_1 \leq x_2$ (when $\Delta > 0$ ) and $a > 0$ : $f(x) \leq 0$ for $x \in [x_1, x_2]$ ; $f(x) \geq 0$ for $x \leq x_1$ or $x \geq x_2$ . If $a < 0$ , the signs reverse.

Worked examples

Five examples with increasing difficulty — from direct application of the quadratic formula to profit optimization. Each example cites the source book.

Example— 1· Quadratic formula — complete solution (application)

Problem: Solve $2x^2 - 7x + 3 = 0$ .

Strategy: Identify $a, b, c$ ; compute $\Delta$ ; apply the quadratic formula; verify with Vieta's formulas.

Solution:

Coefficients. $a = 2$ , $b = -7$ , $c = 3$ .
Discriminant. $\Delta = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25 > 0$ . Two distinct real roots.
Quadratic formula. $x = \dfrac{7 \pm \sqrt{25}}{4} = \dfrac{7 \pm 5}{4}$ $x = \frac{7 \pm 25}{4} = \frac{7 \pm 5}{4}$ .
- $x_1 = \dfrac{7 + 5}{4} = 3$
- $x_2 = \dfrac{7 - 5}{4} = \dfrac{1}{2}$
Factoring. $2(x - 3)(x - \tfrac{1}{2}) = (x - 3)(2x - 1)$ .

Verification (Vieta): $x_1 + x_2 = 3 + \tfrac{1}{2} = \tfrac{7}{2} = -b/a = 7/2$ ✓ and $x_1 \cdot x_2 = 3/2 = c/a = 3/2$ ✓.

Source. OpenStax — College Algebra 2e §5.3 — license CC-BY 4.0.

Example— 2· Vertex, vertex form, and concavity (intermediate)

Problem: For $f(x) = -3x^2 + 12x - 7$ , determine the vertex, vertex form, concavity, and maximum value.

Strategy: Use $x_V = -b/(2a)$ , compute $y_V = f(x_V)$ , write the vertex form; concavity comes from the sign of $a$ .

Solution:

$x_V$ . $x_V = -12/(2 \cdot (-3)) = -12/(-6) = 2$ .
$y_V$ . $f(2) = -3(4) + 12(2) - 7 = -12 + 24 - 7 = 5$ .
Vertex form. $f(x) = -3(x - 2)^2 + 5$ .
Concavity. $a = -3 < 0$ : parabola opens downward. Vertex is a maximum.
Maximum value of $f$ : $y_V = 5$ , attained at $x = 2$ .

Verification: Expand $-3(x-2)^2 + 5 = -3(x^2 - 4x + 4) + 5 = -3x^2 + 12x - 12 + 5 = -3x^2 + 12x - 7$ ✓.

Source. Stitz–Zeager — Precalculus §2.3 — license CC-BY-NC-SA.

Example— 3· Quadratic inequality with sign analysis (intermediate)

Problem: Solve $x^2 - 2x - 8 \leq 0$ .

Strategy: Find the roots of the associated equation; use the concavity to determine where the parabola lies below the $x$ -axis.

Solution:

Roots. $x^2 - 2x - 8 = 0$ : $\Delta = 4 + 32 = 36$ . $x = (2 \pm 6)/2$ : $x_1 = -2$ , $x_2 = 4$ .
Factoring. $(x + 2)(x - 4) = 0$ .
Sign analysis. $a = 1 > 0$ : "smiling" parabola. It lies below the axis ( $f \leq 0$ ) between the roots, including the endpoints (non-strict inequality).
Solution. $x \in [-2, 4]$ .

Verification:

At $x = 0$ (interior): $0 - 0 - 8 = -8 \leq 0$ ✓.
At $x = 5$ (exterior): $25 - 10 - 8 = 7 > 0$ ✓ (not in solution).
At $x = -2$ (boundary): $4 + 4 - 8 = 0 \leq 0$ ✓ (included).

Source. OpenStax — College Algebra 2e §5.3 — license CC-BY 4.0.

Example— 4· Area optimization with rectangular fence (advanced)

Problem: A farmer has 120 m of fencing and wants to enclose a rectangular pasture against an existing wall (the fence covers only 3 sides). What dimensions maximize the area?

Strategy: Model the area as a function of a single variable using the fence-length constraint. The vertex of the resulting quadratic gives the maximum.

Solution:

Variables. Let $x$ be the depth (sides perpendicular to the wall) and $y$ the width parallel to the wall. Fence: $2x + y = 120 \Rightarrow y = 120 - 2x$ .
Area. $A(x) = x \cdot y = x(120 - 2x) = 120x - 2x^2$ .
Physical domain. $x > 0$ and $y > 0 \Rightarrow x \in (0, 60)$ .
Vertex. $a = -2 < 0$ — has a maximum. $x_V = -120/(2 \cdot (-2)) = 30$ m. $y = 120 - 60 = 60$ m.
Maximum area. $A(30) = 30 \cdot 60 = 1800$ m².

Verification: $A(25) = 25 \cdot 70 = 1750 < 1800$ ✓ and $A(35) = 35 \cdot 50 = 1750 < 1800$ ✓.

Remark. The optimal ratio is $y = 2x$ (the dimension parallel to the wall is twice the perpendicular one). This generalizes: with $n$ sides fenced, the optimum divides the available length equally among the fenced sides.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 6 — free license.

Example— 5· Monopolist profit maximization (real modeling)

Problem: A firm has revenue $R(q) = 180q$ (R$/unit, fixed price) and cost $C(q) = 2q^2 + 20q + 500$ . Determine the quantity that maximizes profit and compute the maximum profit.

Strategy: Define $L(q) = R(q) - C(q)$ ; identify it as a quadratic; use the vertex.

Solution:

Profit. $L(q) = 180q - (2q^2 + 20q + 500) = -2q^2 + 160q - 500$ .
Concavity. $a = -2 < 0$ : has a maximum.
$q^*$ . $q_V = -160/(2 \cdot (-2)) = 160/4 = 40$ units.
Maximum profit. $L(40) = -2(1600) + 160(40) - 500 = -3200 + 6400 - 500 = 2700$ reais.
Verification: $L(39) = -2(1521) + 6240 - 500 = -3042 + 5740 = 2698 < 2700$ ✓.

Interpretation (MR = MC): Marginal revenue $R'(q) = 180$ ; marginal cost $C'(q) = 4q + 20$ . Setting equal: $4q + 20 = 180 \Rightarrow q = 40$ — the vertex of the quadratic coincides with the microeconomic optimum.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 6 — free license.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 26Understanding 8Modeling 4Challenge 2

Sources

Only books that directly fed the text and exercises. Full catalog at /livros.

OpenStax College Algebra 2e — Jay Abramson et al. · 2022 · EN · CC-BY 4.0 · §5.1–5.3 (quadratic functions, vertex, discriminant, optimization). Primary source for Blocks A, C, and D.
Stitz–Zeager — Precalculus — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §2.3 (vertex form, Vieta, transformations, challenges). Primary source for Blocks B and E.
Yoshiwara — Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · free · chs. 6–7 (area optimization, profit, ballistics). Primary source for Block D and the practical door.