Lesson 10: Introduction to Quadratics

A quadratic function has the form f(x) = ax² + bx + c with a ≠ 0. Its graph is a parabola: a U shape when a > 0 (opening upward, with a lowest point) or an upside-down U when a < 0. Everything in this lesson — vertex, axis, intercepts — is a feature of that curve, and Lesson 9's factoring is what unlocks its solutions.

Anatomy of a parabola

• Vertex: the turning point — the minimum (a > 0) or maximum (a < 0).
• Axis of symmetry: the vertical line through the vertex, x = −b/(2a). The parabola is a mirror image across it.
• y-intercept: the point (0, c), read straight from the constant term.
• Roots (x-intercepts): where the graph crosses the x-axis — the solutions of ax² + bx + c = 0.

Worked example 1: reading a parabola from a table

For y = x² − 4x + 3, evaluate at x = 0 to 4:

The table itself tells the story: the values fall then rise symmetrically around x = 2, where y = −1. So the axis of symmetry is x = 2 (matching −b/(2a) = 4/2 = 2), the vertex is (2, −1), the y-intercept is 3, and the roots are x = 1 and x = 3 — confirmed by factoring: x² − 4x + 3 = (x − 1)(x − 3), and the zero product property gives the same roots. Three views — table, graph, factors — one object.

Worked example 2: a downward parabola

For y = −2x² + 8x − 5, a < 0, so the parabola opens downward and the vertex is a maximum. Axis: x = −8/(2 × −2) = 2. Vertex height: y = −2(4) + 8(2) − 5 = −8 + 16 − 5 = 3, so the vertex is (2, 3) and the function's maximum value is 3. This quadratic has roots too — but they are not whole numbers, which is exactly the gap the quadratic formula fills.

Solving quadratics: three tools

• Factoring (fastest when it works): x² + 2x − 8 = 0 → (x + 4)(x − 2) = 0 → x = −4 or x = 2.
• The quadratic formula (always works): for ax² + bx + c = 0, x = (−b ± √(b² − 4ac)) / (2a). Apply to 2x² + 3x − 2 = 0: the discriminant is 3² − 4(2)(−2) = 9 + 16 = 25, so x = (−3 ± 5)/4, giving x = 1/2 or x = −2. Check x = 1/2: 2(1/4) + 3(1/2) − 2 = 0.5 + 1.5 − 2 = 0 ✓.
• The discriminant predicts the count before you compute: b² − 4ac > 0 → two real roots; = 0 → exactly one (repeated) root; < 0 → no real roots — the parabola never reaches the x-axis. Example: x² + 4x + 5 has discriminant 16 − 20 = −4 < 0, no real roots. Consistently, its vertex is (−2, 1): the lowest point of an upward parabola is still 1 unit *above* the axis, so it can never cross.

Modelling application: a thrown ball

A ball is thrown from a platform 6 feet high, and its height after t seconds is h(t) = −16t² + 32t + 6 feet (the −16t² is gravity's signature in feet-and-seconds problems).

• At t = 1: h = −16 + 32 + 6 = 22 feet.
• The vertex is at t = −32/(2 × −16) = 1 second, so the maximum height is 22 feet, reached 1 second after the throw. (The symmetry is real: the height at t = 0.5 equals the height at t = 1.5 — check: both give 18 feet.)
• The constant 6 is the launch height — the platform — just as c was the y-intercept. Reading a model's constants in context is the same skill as reading the graph's features.

Common mistakes

• The sign in −b/(2a). For y = x² − 4x + 3, b = −4, so −b/(2a) = −(−4)/2 = +2. Students who substitute b as "4" place the axis at −2 and shift the whole graph wrongly. Write a, b, c down *with signs* before using any formula.
• Confusing the vertex with the roots. In Worked example 1 the vertex is (2, −1) and the roots are 1 and 3 — the vertex's x-coordinate (2) is the *midpoint* of the roots, not a root itself. Mixing these up costs points in both directions.
• Forgetting the ±. x = (−3 ± 5)/4 is two computations. Dropping the minus branch discards x = −2 — half the answer, and often the half a follow-up question needs.
• Applying formulas to a non-standard equation. 2x² = 8 − x must first be rearranged to 2x² + x − 8 = 0 (a = 2, b = 1, c = −8). Reading b and c off the unrearranged form gives the wrong discriminant entirely.

Exam Experience

Question 1. What is the vertex of y = x² − 6x + 5? A. (3, −4) B. (−3, −4) C. (3, 4) D. (−3, 32) Answer: A. Axis: x = 6/2 = 3; height: 9 − 18 + 5 = −4. Option B takes the axis sign wrong; C is the vertex of the upside-down twin.

Question 2. The solutions of x² − 9 = 0 are: A. x = 3 only B. x = −3 only C. x = 3 or x = −3 D. x = 9 or x = −9 Answer: C. A difference of squares: (x − 3)(x + 3) = 0. Equations with x² produce pairs of opposite roots here — choosing only 3 throws away the negative crossing.

Question 3. The discriminant of x² + 2x + 1 is 0. This means the equation has: A. Two distinct real roots B. Exactly one (repeated) real root C. No real roots D. Roots 0 and −2 Answer: B. Discriminant zero is the boundary case: the parabola just *touches* the x-axis. Indeed x² + 2x + 1 = (x + 1)², with the single repeated root x = −1 — the vertex sitting exactly on the axis.

Question 4 (constructed response). A rocket's height is modelled by h(t) = −16t² + 48t + 4 feet, where t is seconds after launch. (a) Find the time at which the rocket reaches its maximum height, and that height. (b) Find when the rocket hits the ground (h = 0), using the quadratic formula. Give your answer to two decimal places. (c) The formula also produces a negative time. Explain why it is rejected.

Scoring guide (4 points).

• 1 point: vertex time t = −48/(2 × −16) = 1.5 seconds.
• 1 point: h(1.5) = −16(2.25) + 48(1.5) + 4 = −36 + 72 + 4 = 40 feet.
• 1 point: ground time from −16t² + 48t + 4 = 0: discriminant 48² − 4(−16)(4) = 2304 + 256 = 2560; t = (−48 − √2560)/(−32) form, equivalently t = (48 + √2560)/32 ≈ (48 + 50.60)/32 ≈ 3.08 seconds (accept 3.08–3.09).
• 1 point: the negative root (t ≈ −0.08 s) is rejected because it describes a moment *before launch* — outside the model's time domain — leaving 3.08 s as the physically meaningful answer.

Deeper still: the square-root method and completing the square as a solver

Before the formula, two direct methods deserve fluency. The square-root method handles equations with a lone squared expression: x² = 25 → x = ±5 (remember both signs — the equation asks for *all* numbers whose square is 25); and (x − 3)² = 16 → x − 3 = ±4 → x = 7 or x = −1. This is precisely the root-finding move used on vertex form in Lesson 11 — same algebra, new outfit. Completing the square converts any quadratic into that shape. Solve x² + 6x + 5 = 0: move the constant, x² + 6x = −5; add (6/2)² = 9 to both sides: x² + 6x + 9 = 4, i.e. (x + 3)² = 4; take roots: x + 3 = ±2 → x = −1 or x = −5. Check: 1 − 6 + 5 = 0 ✓; 25 − 30 + 5 = 0 ✓. Factoring agrees: (x + 1)(x + 5). Three methods, one answer set — and notice the vertex peeking through: (x + 3)² = 4 rewrites the parabola as y = (x + 3)² − 4, vertex (−3, −4), whose x-intercepts are exactly the roots found. Algebra and geometry keep certifying each other; that cross-checking is the deepest habit this course teaches.

One more idea: consistency among vertex, table and roots

Close the loop on Worked example 2's y = −2x² + 8x − 5 pattern with a friendlier case: y = 2x² − 4x + 1. Its vertex: x = 4/4 = 1, y = 2 − 4 + 1 = −1 → (1, −1). Its discriminant: 16 − 8 = 8 > 0, promising two roots: x = (4 ± √8)/4 = 1 ± √2/2 ≈ 0.29 and 1.71. Now the audit: the roots' midpoint is (0.29 + 1.71)/2 = 1 — the axis of symmetry, as required — and a value table confirms the shape:

Values at x = 0 and x = 2 match (both 1) around the vertex x = 1 ✓, and the sign changes between 0–1 and 1–2 bracket the two roots ✓. When vertex, discriminant, roots and table all agree, the analysis is certainly right; when any two disagree, one of them has an arithmetic error and the disagreement tells you where to hunt. Build this cross-checking reflex and quadratics stop being four separate topics — they become one object inspected four ways.