Quadratic Functions: Graphs, Zeros, and the Vertex
Overview
This is a self-contained, low-tech substitute packet in which students practice quadratic functions. Working alone with a pencil (a graphing or scientific calculator is allowed but not required), students evaluate a quadratic and factor a trinomial, study a worked example that shows one quadratic in standard form y = ax² + bx + c as a table and a labeled parabola, and then apply the ideas: reading the zeros (roots), vertex, axis of symmetry, and y-intercept from a graph; factoring a simple quadratic to find roots; matching a quadratic to a projectile context and interpreting the vertex as a maximum; and fixing a sign error in factoring. On a block schedule they also read vertex form y = a(x − h)² + k and model a maximum-area fence problem. They finish by justifying whether a classmate correctly located a vertex, using a claim, evidence, and reasoning. A key idea throughout is telling a root (where y = 0) apart from the vertex (the turning point, halfway between the roots).
At a glance
Course: Algebra II (Grades 9–12)
Subject: Mathematics
Time: about 50 minutes on a standard period, or about 90 minutes on a block period (block adds vertex form and a maximum-area modeling task)
Materials: printed packet and a pencil; a graphing or scientific calculator is allowed but not required (no computer or internet)
Work mode: independent
Standards (provisional): 19 TAC §111.40 (Algebra II) — graphing quadratic functions and identifying key attributes (vertex, axis of symmetry, zeros/x-intercepts, y-intercept, maximum/minimum); solving quadratic equations by factoring; and interpreting the meaning of the vertex and intercepts in real-world contexts. Provisional — pending educator verification against the current official TAC source. Standards are paraphrased, not quoted; this packet is not claimed to be "aligned to the TEKS" until reviewed.
Accessible version of the student activity
The full student activity is reproduced below in plain, screen-reader-friendly HTML. It reflows on phones and at 200% zoom. Write your answers on the printed packet, and show your work.
Start (6 minutes) — Retrieval warm-up
- Evaluate a quadratic: f(x) = x² − 4x + 3. Find f(0) and f(2). Show the squaring, the multiplication, and the addition. (Square the input first, then multiply, then combine.)
- Factor x² + 5x + 6 into two binomials (two numbers that multiply to 6 and add to 5), then find the two values of x that make the product 0 (Zero Product Property).
Build (8–12 minutes) — Study the math
A quadratic function graphs as a parabola (a symmetric U-shaped curve). Its standard form is y = ax² + bx + c (a ≠ 0). If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum. The number c is the y-intercept (0, c). The vertex is the turning point; its x-coordinate is x = −b ÷ (2a), and substituting it back gives the y-coordinate. The axis of symmetry is the vertical line x = −b ÷ (2a) through the vertex; the parabola is a mirror image across it. The zeros (roots or x-intercepts) are the x-values where y = 0. To find them by factoring, set ax² + bx + c = 0, factor to (x − r₁)(x − r₂) = 0, and use the Zero Product Property (if a product is 0, a factor is 0) to get x = r₁ or x = r₂. In a real story the vertex is often a maximum (greatest height, largest area) or a minimum (least cost). Remember: a root is where the curve crosses the x-axis, while the vertex is where the curve turns — halfway between the two roots.
Worked example — y = x² − 4x + 3
Here a = 1, b = −4, c = 3. Because a = 1 > 0, the parabola opens up, so the vertex is a minimum. The y-intercept is c = 3, the point (0, 3). Axis of symmetry / vertex x: x = −b ÷ (2a) = −(−4) ÷ (2·1) = 4 ÷ 2 = 2, so the axis is the line x = 2. Vertex y: f(2) = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1, so the vertex is (2, −1). Zeros by factoring: x² − 4x + 3 = (x − 1)(x − 3) = 0, so x = 1 or x = 3 — the parabola crosses the x-axis at (1, 0) and (3, 0). The zeros 1 and 3 are the same distance (1 unit) on each side of the axis x = 2, and their midpoint (1 + 3) ÷ 2 = 2 is the axis.
| x | f(x) | Note |
|---|---|---|
| 0 | 3 | y-intercept (0, 3) |
| 1 | 0 | zero / x-intercept |
| 2 | −1 | vertex (minimum) |
| 3 | 0 | zero / x-intercept |
| 4 | 3 | mirror of (0, 3) |
- Using f(x) = x² − 4x + 3, find f(4), and explain in one sentence why f(4) equals f(0) (use the axis of symmetry x = 2).
Apply (20–26 minutes) — Use what you know
Item 4 uses a parabola (opens up) through the points (−1, 0), (0, −3), (1, −4), (2, −3), and (3, 0).
| x | y |
|---|---|
| −1 | 0 |
| 0 | −3 |
| 1 | −4 |
| 2 | −3 |
| 3 | 0 |
- Read the graph/table above:
- 4a. Name the two zeros (x-intercepts) where y = 0.
- 4b. Give the vertex (lowest point) as an ordered pair, and state whether it is a maximum or a minimum.
- 4c. Give the axis of symmetry as x = ___, and explain how the two zeros show the axis.
- 4d. Give the y-intercept (the value of y when x = 0).
- Factor to find the roots: solve x² − 7x + 10 = 0 by factoring; show the two numbers (multiply to 10, add to −7), the factored form, and each root, then state the axis of symmetry (halfway between the roots).
- Match a quadratic to a context: a soccer ball's height is h(t) = −16t² + 32t (a = −16, b = 32, c = 0).
- 6a. Does the parabola open up or down, and is the vertex a maximum or minimum? Use the sign of a.
- 6b. Find the time of the vertex with t = −b ÷ (2a), then the greatest height by evaluating h there; show your steps and interpret the vertex (when highest, how high).
- Error analysis (a sign error in factoring): a student wrote "x² − x − 6 = (x − 2)(x − 3) = 0, so x = 2 or x = 3." Find the mistake, explain it, and give the correct roots.
Block only (~12–15 minutes) — Extra applied task
- Vertex form and a second model. Vertex form is y = a(x − h)² + k, with vertex (h, k) and axis x = h.
- 8a. For y = 2(x − 3)² − 5, write the vertex and the axis of symmetry; state whether it opens up or down and whether the vertex is a max or min.
- 8b. Area modeling: with 40 ft of fence against a barn wall (three sides), width x gives area A(x) = x(40 − 2x) = −2x² + 40x. Find the width x at the vertex (x = −b ÷ (2a)) and the maximum area; show your steps and interpret the vertex (best width, biggest area).
Explain (6–10 minutes) — Justify your strategy (Claim, Evidence, Reasoning)
Question 9. A classmate says, "For y = x² − 6x + 8, the vertex is at x = 4 because the roots are 2 and 4." Is the classmate right? Find the true roots (by factoring) and the true vertex (x = −b ÷ (2a)), and decide. Write a Claim (your answer), Evidence (the roots and the vertex x), and Reasoning (why the axis of symmetry sits halfway between the roots, not at a root).
Sentence stems you may use: "The classmate is ___ because…"; "Factoring x² − 6x + 8 gives… so the roots are…"; "The axis of symmetry is x = −b ÷ (2a) = …"; "The vertex sits halfway between the roots because…"; "A root is where y = 0, but the vertex is where the curve turns, so…".
Close (5 minutes) — ACE
- Articulate: explain the difference between a zero (root) and the vertex. Which is where the curve crosses the x-axis, and which is where it turns?
- Connect: name one packet item where you found a vertex and used it as a maximum or minimum in a real context.
- Extend: give a new real-life example where finding a vertex (a maximum or minimum) answers a useful question, not used in this packet.
Continue (optional) — Early finisher
Build your own quadratic: choose two whole-number roots, write the factored form (x − r₁)(x − r₂), multiply it into standard form y = x² + bx + c, find the axis of symmetry and the vertex, and make a small table of 5 outputs and sketch the parabola, marking the vertex, the axis, and both zeros. Check that your roots are the same distance on each side of your axis.
Turn in
Hand in the whole packet with your name, class period, and date, with items 1–7, 9, and the ACE box answered and your work shown (item 8 too if you are on a block schedule). Include the optional challenge if you did it.