Week 9 · Lecture outline

Week 9 — Lecture Outline · Quadratic Functions & Their Graphs

College Algebra · MATH 120 Fall 2026 · Prof. Calloway Fictional sample

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 6 — Analyze quadratic functions by identifying vertex, axis of symmetry, intercepts, and maximum/minimum values, and graph parabolas in both standard and vertex form.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/graphical representations and interpret in context)
Meeting pattern: 2 sessions × 75 min = 150 min. Segment minutes below total ~150; scale to your own pattern.

Week at a Glance


The week's big question	"Given a quadratic function, what can I read off its graph — and how do I find each feature from the formula?"
By the end of the week, students can…	(1) identify the vertex and axis of symmetry from both vertex form and standard form; (2) find the y-intercept and x-intercepts and determine how many intercepts exist using the discriminant; (3) decide whether a parabola opens up or down and identify the minimum or maximum value; (4) apply a quadratic function to a real-world optimization or projectile problem, finding and interpreting the vertex.
Key vocabulary	parabola, vertex, axis of symmetry, vertex form, standard form, opening direction, minimum value, maximum value, y-intercept, x-intercepts (zeros), discriminant
Materials	slides (Deck 9), the week's readings + video links, Desmos (or GeoGebra) for graphing and checking, one approved chatbot (Gemini / Claude / ChatGPT) for the AI-critique moment and the tutorial
Timing note	8 segments, ~150 min total. Session 1 = Segments 1–4 (~73 min). Session 2 = Segments 5–8 (~77 min).

Segment 1 — Hook & the Promise (8 min) · Session 1 opens

Hook. "Quick show of hands: who's ever watched a basketball shot arc through the air, or a water fountain curve, or a ball thrown at the park?" (Pause.) "Every single one of those paths is a parabola — the graph of a quadratic function. And the highest point of each arc? That's a vertex. Today we're going to learn to read that vertex — and everything else about a parabola — straight from its equation."

The promise (write it on the board): "By the end of this week you can look at a quadratic function and immediately say: here's the vertex, here's the axis of symmetry, here's whether it has a maximum or minimum, and here's where it crosses the axes — without plotting a single point."

Why it matters line (memory hook): "Quadratic functions are the simplest nonlinear functions. Every curve you'll meet the rest of this course, and in calculus, builds on the ideas we lock down this week."

Welcome-back framing: remind students Week 7 taught solving quadratic equations (finding x when y = 0). This week we keep the entire function and ask: what is its shape, and what does that shape tell us?

Segment 2 — The Parabola & Vertex Form (22 min)

Plain language first. The graph of any quadratic function f(x) = ax² + bx + c is a parabola — a symmetric U-shape (or inverted U). Every parabola has exactly one turning point: the vertex. The vertical line through the vertex is the axis of symmetry — fold the parabola along that line and the two halves match perfectly.

Vertex form — the direct read:

f(x) = a(x − h)² + k
- Vertex: (h, k) — the horizontal shift h and the vertical shift k are read directly from the formula.
- Axis of symmetry: x = h.
- Direction: a > 0 → parabola opens up (like a cup); a < 0 → opens down (like a hill).
- Minimum value (if a > 0): the y-coordinate of the vertex = k.
- Maximum value (if a < 0): the y-coordinate of the vertex = k.

Memory hook: "In vertex form, the vertex is (h, k) — read h and k straight out of the formula. The only trick: the formula has (x − h), so if you see (x − 2)², h = 2, not −2."

One fully worked example (every step out loud):

Analyze and sketch f(x) = (x − 2)² + 3.
1. Vertex: h = 2, k = 3 → vertex is (2, 3).
2. Axis of symmetry: x = 2.
3. Direction: a = 1 > 0 → opens up → minimum at vertex.
4. Minimum value: 3 (the y-coordinate of the vertex).
5. y-intercept: f(0) = (0 − 2)² + 3 = 4 + 3 = 7 → point (0, 7).
6. x-intercepts: set (x − 2)² + 3 = 0 → (x − 2)² = −3 → no real solution (a perfect square can't be negative) → no x-intercepts.
7. Sketch: vertex at (2, 3), symmetric about x = 2, passes through (0, 7) and by symmetry (4, 7), opens up.

The sign trap — name it loudly:

❌ "I see (x − 2)² + 3, so h = −2."
✅ Cure: the vertex form has (x − h), so we subtract h. Since the formula shows x − 2, we have h = +2. The vertex is (2, 3). If the formula showed (x + 2)² = (x − (−2))², then h = −2.

Segment 3 — Standard Form & the Vertex Formula (25 min)

Hook back in: "Vertex form is clean, but most quadratics you'll meet are in standard form: f(x) = ax² + bx + c. Good news: there's a formula that finds the vertex from the coefficients in about five seconds."

Plain language first. Standard form f(x) = ax² + bx + c:
- y-intercept: set x = 0 → f(0) = c → point (0, c). It's just c.
- Vertex x-coordinate: x = −b/(2a) (derived by completing the square — see Segment 4 for the connection).
- Vertex y-coordinate: evaluate f at that x.
- Axis of symmetry: x = −b/(2a) (same as the vertex x-coordinate).
- Direction & min/max: same rule — sign of a.

Memory hook: "The vertex is where the slope is zero — and for a quadratic, that's exactly at x = −b/(2a). Memorize it as a unit: 'negative b over two a'."

One fully worked example (every step):

Analyze f(x) = x² − 6x + 5.
1. Identify: a = 1, b = −6, c = 5.
2. Vertex x: x = −(−6)/(2·1) = 6/2 = 3.
3. Vertex y: f(3) = 9 − 18 + 5 = −4 → vertex is (3, −4).
4. Axis of symmetry: x = 3.
5. Direction: a = 1 > 0 → opens up → minimum at vertex.
6. Minimum value: −4.
7. y-intercept: f(0) = 0 − 0 + 5 = 5 → point (0, 5).
8. x-intercepts: x² − 6x + 5 = 0 → (x − 1)(x − 5) = 0 → x = 1 and x = 5 → points (1, 0) and (5, 0).
9. Sketch: vertex (3, −4), axis x = 3, opens up, crosses x-axis at 1 and 5, y-intercept 5.

Another quick example (opens down — maximum):

f(x) = −x² + 4x + 1. a = −1 < 0 → opens down → maximum.
Vertex: x = −4/(2·(−1)) = −4/−2 = 2. f(2) = −4 + 8 + 1 = 5. Vertex (2, 5). Maximum value = 5.

Segment 4 — Misconceptions & Cures (18 min) · Session 1 closes (~73 min)

Name each misconception out loud, then cure it:

❌ "I see (x − 2)² + 3, so the vertex is at (−2, 3)."
✅ Cure: the subtraction is already in the formula. Vertex form is f(x) = a(x − h)² + k, and h is what you subtract FROM x. Since we have x − 2, h = 2. The vertex is (2, 3). Say it: "If it's (x minus h), then h is positive; if it's (x plus 2) = (x minus (−2)), then h is negative."
❌ "The axis of symmetry is the point (3, −4), not the line x = 3."
✅ Cure: the axis is a vertical line, written x = (a number). The vertex is a point — both coordinates. They share the same x-value but are different things.
❌ "For x = −b/(2a), I just compute b/(2a)." (forgetting the negative)
✅ Cure: for f(x) = x² − 6x + 5, b = −6. Without the negative: (−6)/(2) = −3. Wrong. With the negative: −(−6)/2 = 6/2 = 3. Correct. The formula has a built-in minus sign; don't drop it.
❌ "f(x) = −x² + 4x + 1 has a minimum at the vertex because the vertex y-value is 5."
✅ Cure: the direction is determined by a, not by the vertex y-value. Here a = −1 < 0 → the parabola opens down → the vertex is the maximum, not the minimum. It's the highest point, not the lowest.

Interaction — Quick Vote (10 min):
Put four functions on the board; students hold up M (minimum) or X (maximum):
1. f(x) = 3x² − 2x + 1 (M, a = 3 > 0)
2. f(x) = −(x − 4)² + 7 (X, a = −1 < 0)
3. f(x) = x² + 6x − 3 (M, a = 1 > 0)
4. f(x) = −2x² + 8x (X, a = −2 < 0)
Debrief: the sign of a is the only thing that matters. The size of a controls how wide or narrow the parabola is, not its direction.

Segment 5 — Intercepts & the Discriminant (20 min) · Session 2 opens

Hook back in: "Last session: we found the vertex and axis of symmetry. Today we finish the picture: where does the parabola cross the axes?"

y-intercept: always easy — plug in x = 0.

For f(x) = ax² + bx + c: f(0) = c. The y-intercept is always (0, c).

x-intercepts (zeros): set f(x) = 0 and solve. Can use factoring, the quadratic formula, or vertex form (if a>0 and k<0, there are two; if k=0, one; if k>0, none for a>0).

Discriminant shortcut (no solving needed):

For ax² + bx + c = 0, the discriminant is b² − 4ac.
- b² − 4ac > 0 → two distinct real x-intercepts.
- b² − 4ac = 0 → one real x-intercept (vertex touches the x-axis).
- b² − 4ac < 0 → no real x-intercepts (parabola floats entirely above or below the x-axis).

Memory hook: "Discriminant tells you how many times the parabola touches the x-axis — positive = two, zero = one, negative = none. You don't need to solve to know."

One worked example (using discriminant):

How many x-intercepts does y = x² + 1 have?
Discriminant = 0² − 4(1)(1) = −4 < 0 → no real x-intercepts. (The parabola sits entirely above the x-axis — you can see this because the vertex is (0, 1) and it opens up.)

Worked example connecting everything:

f(x) = x² − 6x + 5 (from Segment 3):
- Discriminant: (−6)² − 4(1)(5) = 36 − 20 = 16 > 0 → two x-intercepts ✓ (we found 1 and 5).
- y-intercept: (0, 5) → equal to c = 5 ✓.

Segment 6 — Converting Between Forms + Technology (20 min)

Connecting the two forms: completing the square converts standard → vertex form, and expanding converts vertex → standard. You don't need to complete the square on the quiz — the vertex formula −b/(2a) is faster — but recognizing the connection is conceptually important.

Example (expand vertex form to get standard):

f(x) = 2(x − 1)² − 3 = 2(x² − 2x + 1) − 3 = 2x² − 4x + 2 − 3 = 2x² − 4x − 1.
Check: a = 2, b = −4 → vertex x = −(−4)/(4) = 1. f(1) = 2 − 4 − 1 = −3. ✓ Matches.

Technology workflow — sketch and verify in Desmos (exact steps):
1. Open desmos.com/calculator (free, no login).
2. Type f(x) = x^2 - 6x + 5 on line 1.
3. Click the vertex label that Desmos shows: it should say (3, −4). That's your answer.
4. Click the x-axis intercepts: (1, 0) and (5, 0). ✓
5. Click the y-intercept: (0, 5). ✓
6. Now type g(x) = (x-3)^2 - 4 on line 2 — same graph, confirming the two forms are equivalent.

Desmos is your check, not your solution. On the quiz, show the algebra — then use Desmos to verify.

Segment 7 — Applications: Max/Min Word Problems (20 min)

Plain language first. In any max/min application with a quadratic model, the answer lives at the vertex. The setup is always:
1. Write a quadratic function modeling the situation.
2. Identify a, b, c.
3. Find the vertex using x = −b/(2a) → interpret as the input at the optimal value.
4. Evaluate f at the vertex → interpret as the optimal output value.

Worked example — projectile motion:

A ball is launched from a 80-foot platform. Its height (in feet) after t seconds is h(t) = −16t² + 64t + 80.
1. a = −16, b = 64, c = 80. a < 0 → opens down → maximum height.
2. Time at maximum: t = −64/(2·(−16)) = −64/(−32) = 2 seconds.
3. Maximum height: h(2) = −16(4) + 64(2) + 80 = −64 + 128 + 80 = 144 feet.
4. Interpret: the ball reaches its maximum height of 144 feet at t = 2 seconds.

Key scaffolding: write the interpretation sentence every time. "The maximum height is 144 feet, reached 2 seconds after launch" is the answer, not just the number 144.

Segment 8 — AI-Critique, Callback & Hand-off (12 min) · Session 2 closes (~77 min)

AI-critique moment (students verify, not consume):

Paste this to an approved chatbot: "Find the vertex of f(x) = (x + 3)² − 5 and state whether it has a maximum or minimum."
Then check its work by hand. Common errors:
- Chatbots sometimes report the vertex as (−3, −5) ✓ but then wrongly say "maximum" (they forget to check the sign of a, which is +1, so it's a minimum).
- Some will say vertex = (+3, −5) — reading (x + 3) as h = 3 instead of h = −3.
Correct answer: vertex (−3, −5), minimum (a = 1 > 0).
Your job all semester: the tool drafts, you judge.

Callback + tease:
- Callback: "Everything this week sits on top of Week 7 — you already know how to solve quadratic equations. This week we kept the function alive and studied its shape. The algebraic moves are the same; the picture is new."
- Tease next week: "Week 9 is the bridge from quadratics to the world beyond. In Week 10 we generalize — polynomials of degree 3, 4, and higher — and ask the same questions: where are the zeros, which way does it go at the ends, what's the shape?"

Hand-off (the week's graded work):
- Lecture Tutorial 9 (AI tutor, share-link submission) — vertex form, standard form, vertex formula, intercepts, direction, min/max.
- Quiz 9 (end of week, no AI) and Discussion 9 ("Parabolas in Your World").
- Assignment 9 — AI-coached, self-scored.

Instructor FAQ — Common Stumbles

Student says / does	Quick cure
Reads vertex of (x − 2)² + 3 as (−2, 3).	The formula has (x − h), so h = +2. Vertex is (2, 3). The subtraction is already built in.
Uses b/(2a) instead of −b/(2a).	The formula is negative b over 2a. For f = x² − 6x + 5, b = −6: without the minus you get (−6)/2 = −3; with the minus you get 6/2 = 3. Always write the minus first.
Says "the axis of symmetry is the vertex."	Axis = a line, x = h. Vertex = a point, (h, k). They share the x-value h but are different objects.
Calls the vertex of −x² + 4x + 1 a "minimum" because it's the "turning point."	a = −1 < 0 → opens down → the vertex is the highest point → maximum. Check the sign of a, not the appearance of the vertex coordinates.
Gets the vertex right but forgets to interpret in context (word problems).	Require the interpretation sentence: "The maximum height is ___ feet at t = ___ seconds." A number without units and context is not a complete answer.
Thinks x-intercepts always exist.	Use the discriminant: if b² − 4ac < 0, there are no real x-intercepts. Check with y = x² + 1 (disc = −4).
Confuses y-intercept with vertex.	y-intercept = f(0) = c, always on the y-axis at (0, c). Vertex = the turning point at x = −b/(2a), usually not on the y-axis unless b = 0.

Scope flag

This outline covers Objective 6's graph-analysis component. Completing the square as a derivation is mentioned in Segment 6 but not drilled — students use x = −b/(2a) on all quiz/assignment items. The discriminant is treated as a sign-reader only (no complex roots). Graphing by table-of-values is not taught — Desmos handles visualization, and the algebraic features (vertex, intercepts) anchor the analysis.

~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com