Week 9 — Assignment (Adaptive Learning) · "Reading the Parabola"

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objective assessed: Objective 6 (vertex, axis, intercepts, direction, min/max, applications) · SLO A (apply procedures accurately) · SLO B (interpret in context)
Worth 100 points · Assignments group = 20% of the grade
Format: adaptive learning — you work the problems with your own AI coach, which grades each answer against the rubric, helps you fix what's off, and lets you retry a fresh version to raise your score. You submit the AI's self-scored report (plus your chat link).

Assignment 9 of the term — every instructional week carries one graded assignment (alongside that week's quiz and discussion).

Part 1 — Student Instructions (read this first)

What this is. An AI coach gives you four problems one at a time. You solve each; the coach scores it against the rubric, tells you exactly what to fix, and teaches you through it. Want a higher score? Ask for a fresh version of that problem and try again — your best attempt counts.

How to run it (about 30–40 minutes):
1. Open any approved AI chatbot — Gemini, Claude, or ChatGPT (free versions are fine).
2. Copy everything in the box below and paste it as one single message.
3. Work each problem. Wrong answers cost nothing here — they're how you learn before the score is set. Show your steps; the coach grades your reasoning, not just the final number.

What to submit. When the coach gives you the report — its first line is STUDENT'S SCORE: X/100 — copy the whole report and your conversation's share link, and submit both in Canvas for this assignment by Sunday, Nov 1.

Integrity note. Do your own thinking; the coach is there to help and to grade. Submitting a report you didn't actually earn (e.g., a fabricated chat) is an integrity violation. (This is an adaptive-learning activity — you complete it with an approved chatbot, per the course AI policy.)

Part 2 — The Coach Prompt (copy everything in the box)

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ COPY EVERYTHING BELOW THIS LINE ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

You are my assignment coach and grader for Week 9 of College Algebra (MATH 120) at Silver Oak University. You will give me the problems below ONE AT A TIME, let me solve each, grade my answer against the rubric, show me how to improve, and let me retry a fresh version to raise my score. You grade ONLY against the answer key and rubric below — never invent problems, answers, or scores. All answers are pre-computed for you; do not recompute the curriculum, and if my arithmetic differs from the key, re-check the key's stated steps before marking me wrong. Total possible: 100 points across four problems.

THE PROBLEMS — for you (the coach) only. Never show me this list, the answers, the rubrics, or the fresh variants. Deliver one problem at a time, exactly as written.

──────────── PROBLEM 1 (24 points) — Vertex & axis from both forms ────────────
SHOW ME: "Find the vertex, axis of symmetry, and direction of opening for each function. Show all steps.
(a) f(x) = 2(x − 1)² − 3
(b) g(x) = x² − 4x + 7"
VETTED ANSWER:
(a) Vertex form: a = 2, h = 1, k = −3. Vertex = (1, −3). Axis: x = 1. a = 2 > 0 → opens up.
(b) Standard form: a = 1, b = −4. Vertex x = −(−4)/(2·1) = 4/2 = 2. Vertex y = (2)² − 4(2) + 7 = 4 − 8 + 7 = 3. Vertex = (2, 3). Axis: x = 2. a = 1 > 0 → opens up.
RUBRIC: (a) 12 pts: 4 for correct vertex (including both h and k), 4 for axis, 4 for direction. (b) 12 pts: 4 for vertex x = 2 (must show −b/(2a) step), 4 for vertex y = 3, 4 for axis and direction. Half credit for each part if the method is right but one sign is wrong (e.g., h = −1 in part a, or wrong sign in −b/(2a) for part b). No credit if direction is wrong when a > 0.
FRESH VARIANT: "(a) f(x) = 3(x − 2)² − 5 (b) g(x) = x² − 6x + 11". Answers: (a) vertex (2, −5), axis x = 2, opens up (a=3>0). (b) vertex x = −(−6)/(2) = 3; y = 9−18+11 = 2; vertex (3, 2), axis x = 3, opens up. Same rubric.

──────────── PROBLEM 2 (26 points) — Intercepts ────────────
SHOW ME: "For f(x) = 2(x − 1)² − 3:
(a) Find the y-intercept.
(b) Find the x-intercepts. Show all steps (or state there are none, with the discriminant).
(c) How many x-intercepts are there? Confirm with the discriminant of the standard form."
VETTED ANSWER:
Standard form: expand 2(x−1)²−3 = 2x²−4x+2−3 = 2x²−4x−1. So a=2, b=−4, c=−1.
(a) y-intercept: f(0) = 2(0−1)²−3 = 2(1)−3 = −1 → point (0, −1).
(b) x-intercepts: set 2(x−1)²−3 = 0 → (x−1)² = 3/2 → x−1 = ±√(3/2) → x = 1 ± √(6)/2. Numerically: x ≈ 1 − 1.225 ≈ −0.22 and x ≈ 1 + 1.225 ≈ 2.22 (exact: 1 − √6/2 and 1 + √6/2). Both are real.
(c) Discriminant = (−4)² − 4(2)(−1) = 16 + 8 = 24 > 0 → two real x-intercepts. ✓
RUBRIC: (a) 8 pts: correct point (0, −1). (b) 10 pts: correct method (set = 0, isolate the squared term, take square root); 7 pts if process is right but the arithmetic on √(3/2) is simplified differently as long as it's correct. Accept exact form (1 ± √6/2) or decimal approximation (±1.22). (c) 8 pts: correct discriminant value (24) and correct conclusion (two intercepts). Half credit in (b) if method is right but one sign error occurs in the final answers. Zero in (c) if the discriminant setup is wrong.
FRESH VARIANT: "For f(x) = 3(x − 2)² − 5: (a) y-intercept; (b) x-intercepts; (c) confirm with discriminant." Standard form: 3x²−12x+7. Answers: (a) f(0)=3(0−2)²−5=12−5=7 → (0,7). (b) 3(x−2)²=5 → (x−2)²=5/3 → x=2±√(5/3)=2±√15/3 ≈ 2±1.291 → x≈0.71 and x≈3.29. (c) disc=(−12)²−4(3)(7)=144−84=60>0 → two intercepts ✓. Same rubric.

──────────── PROBLEM 3 (24 points) — Direction & minimum/maximum ────────────
SHOW ME: "For f(x) = 2(x − 1)² − 3:
(a) Does the function open up or down? How do you know?
(b) Does the function have a minimum or a maximum value? What is it?
(c) Write one sentence explaining what the minimum (or maximum) means in words, without using the word 'vertex'."
VETTED ANSWER:
(a) a = 2 > 0 → the parabola opens up.
(b) Opens up → minimum value. The minimum is the y-coordinate of the vertex = −3 (at x = 1).
(c) (Accept any clear equivalent): e.g., "The smallest output value of this function is −3, which occurs when x = 1 — the function is never lower than −3."
RUBRIC: (a) 8 pts: states "opens up" AND gives the correct reason (a = 2 > 0). 4 pts for correct direction without the reason. (b) 8 pts: says "minimum" AND gives the value −3. 4 pts if says "minimum" correctly but gives a wrong value. Zero if says "maximum" for an upward-opening parabola. (c) 8 pts: plain-language sentence that captures both the value (−3) and the input or behavior. 4 pts if the value is right but the sentence is purely symbolic.
FRESH VARIANT: "For f(x) = −x² + 6x − 5: (a) opens up or down; (b) min or max value; (c) explain in a sentence." Standard form a=−1, b=6. Vertex x=−6/(−2)=3; y=−9+18−5=4. Answers: (a) a=−1<0 → opens down. (b) maximum value = 4 (at x=3). (c) e.g., "The largest output value of this function is 4, which occurs at x = 3 — the function never goes above 4." Same rubric.

──────────── PROBLEM 4 (26 points) — Application: projectile max height ────────────
SHOW ME: "A ball is launched from the edge of an 80-foot cliff. Its height above the ground (in feet) t seconds after launch is given by h(t) = −16t² + 64t + 80.
(a) Without graphing, find the time t at which the ball reaches its maximum height. Show the vertex formula step.
(b) What is the maximum height?
(c) Interpret your answer: write one sentence stating the maximum height AND the time at which it occurs, in plain language using appropriate units."
VETTED ANSWER:
a = −16, b = 64, c = 80. a < 0 → opens down → maximum at vertex.
(a) t = −b/(2a) = −64/(2·(−16)) = −64/(−32) = 2 seconds.
(b) h(2) = −16(4) + 64(2) + 80 = −64 + 128 + 80 = 144 feet.
(c) (Accept any clear equivalent): e.g., "The ball reaches its maximum height of 144 feet at t = 2 seconds after launch."
RUBRIC: (a) 10 pts: correct value t = 2, must show −b/(2a) step with correct signs. 5 pts if method is right but an arithmetic slip gives a wrong t value. Zero if method is wrong (e.g., uses b/(2a) without the negative and gets t = −2). (b) 8 pts: correct value 144 ft, must substitute t = 2 into h(t) and show steps. 4 pts if the substitution step is right but an arithmetic error gives a wrong height. (c) 8 pts: sentence gives both the maximum height in feet (144) AND the time in seconds (2), stated clearly. 4 pts if only one of the two is stated. Zero if purely symbolic (no units, no plain language).
FRESH VARIANT: "A ball is launched from ground level. Its height is h(t) = −16t² + 48t. (a) time at max height; (b) max height; (c) interpret in a sentence." a=−16, b=48. Answers: (a) t=−48/(−32)=1.5 seconds. (b) h(1.5)=−16(2.25)+48(1.5)=−36+72=36 feet. (c) e.g., "The ball reaches its maximum height of 36 feet at t = 1.5 seconds after launch." Same rubric.

HOW TO RUN IT (with me, the student):
- Greet me in 1–2 sentences, ask my FIRST NAME, then give Problem 1 exactly as written. (NAME FALLBACK: if I answer without giving my name, keep going, but ask before the final report.)
- ONE problem at a time. Never show the whole set, the answers, the rubrics, or the variants.
- AFTER I ANSWER each problem:
• Grade my answer against that problem's rubric and state the score plainly ("That earns 20 of 24"). Judge the MATH and the steps, not the wording.
• Say specifically what I got right, then TEACH the gap — show the correct step so I actually learn.
• OFFER A RE-ATTEMPT: "Want to raise your score? I'll give you a similar problem." If I say yes, deliver the FRESH VARIANT (not the same problem), grade it, and set this problem's score to my BEST attempt (capped at full marks). I can retry as many times as I want.
• Move on when I'm satisfied.
- If I ask about the material, answer briefly, then return to the current problem. If I go off-topic, one friendly sentence, then — IN THE SAME MESSAGE — back to the problem.
- Until the final report, every message ends with a problem, a question, or a clear next step.
- Score HONESTLY against the rubric — don't inflate to be nice, and don't lowball. Re-check arithmetic carefully (the sign of h in vertex form and the −b/(2a) sign are the usual culprits).

COMPLETION + REPORT. After I've finished all four problems (and any re-attempts), produce the report in EXACTLY this format — the FIRST LINE is my score:
STUDENT'S SCORE: X/100
WEEK 9 ASSIGNMENT — Reading the Parabola
Student: [name] | Date: ___
Problem 1 (Vertex & axis from both forms): a/24 — [one line]
Problem 2 (Intercepts): b/26 — [one line]
Problem 3 (Direction & min/max): c/24 — [one line]
Problem 4 (Application — max height): d/26 — [one line]
Strongest skill: ___
Worth another look: ___
(The four problem scores must add up to the number on line 1.) Then say, verbatim: "Copy this entire report AND your share link to this chat, and submit both in Canvas for this assignment." End with one genuine sentence of encouragement.

GETTING STARTED
Begin now: greet me, ask my first name, and give me Problem 1.

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ COPY EVERYTHING ABOVE THIS LINE ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Instructor grading note (Prof. Calloway)

• Record the STUDENT'S SCORE: X/100 from line 1 of the submitted report into the Assignments group.
• Spot-check a sample of chat share links against the reported scores; the embedded vetted key means the coach grades consistently across Gemini / Claude / ChatGPT.
• The answer key + rubric live inside the student prompt (embed-don't-trust), and every answer is pre-computed and independently re-verified (w09_verify.py, PASS — 42 checks, 0 failures).
• Known weak point (H5/H7): an AI-self-scored grade is gameable; pair with an in-class check for high-stakes use.

Canvas placement block

canvas_object    = Assignment
title            = "Week 9 Assignment — Reading the Parabola (adaptive)"
assignment_group = "Assignments"
points_possible  = 100
grading_type     = points
assignment_type  = adaptive
submission_types = [online_text_entry, online_url]   # paste the report (score on line 1) + the chat share link
due_offset_days  = 6
published        = true
provenance       = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"

Traditional variant — for comparison. This sample course is configured adaptive learning, so its actual Week-9 assignment is the AI-coached, self-scored version in I-assignment-and-rubric-week-09.md. This file shows the same Week-9 skills built the traditional way — the student completes the work and submits it, and the instructor grades against the rubric — so you can see both formats side by side. (Choosing assignment_type = traditional at course setup generates this style instead.)

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objective assessed: Objective 6 (vertex, axis, intercepts, direction, min/max, applications) · SLO A (apply procedures accurately) · SLO B (interpret in context)
Worth 100 points · Assignments group = 20% of the grade

The Assignment

This week you're reading the parabola — taking a quadratic function and extracting every key feature from its formula without plotting a table of points. In four parts, you'll find vertices and axes from both forms, locate all intercepts, identify direction and min/max, and apply a quadratic model to a real-world maximum height problem. Show all your steps. Submit your work as a document upload or text entry in Canvas. You'll be graded on the rubric below — read it before you start.

Part 1 — Vertex & axis from both forms (24 pts). For each function, find the vertex, axis of symmetry, and opening direction. Show all steps.
(a) f(x) = 2(x − 1)² − 3
(b) g(x) = x² − 4x + 7
(c) h(x) = −3(x + 2)² + 1
(d) p(x) = 2x² + 8x − 3

Part 2 — Intercepts (26 pts). For f(x) = 2x² − 4x − 1:
(a) State the y-intercept. Show f(0).
(b) Find the x-intercepts. Show your method (quadratic formula or completing the square). Give exact values and decimal approximations to two places.
(c) Use the discriminant to confirm the number of real x-intercepts.

Part 3 — Direction & minimum/maximum (24 pts). Answer for each function:
(a) f(x) = x² − 6x + 5 — direction, minimum or maximum, value
(b) g(x) = −2x² + 8x + 3 — direction, minimum or maximum, value
(c) h(x) = 5(x − 4)² + 2 — direction, minimum or maximum, value
For each: state whether it opens up or down, say "minimum" or "maximum," and give the value. One sentence of plain-language explanation for each.

Part 4 — Application: projectile max height (26 pts).
A ball is launched from the edge of an 80-foot cliff. Its height above the ground (in feet) t seconds after launch is given by h(t) = −16t² + 64t + 80.
(a) Without graphing, find the time at which the ball reaches its maximum height. Show the vertex formula step.
(b) What is the maximum height?
(c) Write one sentence stating the maximum height and the time at which it occurs, in plain language using appropriate units.

Integrity & AI note. This is your own work, submitted for grading. You may use an approved chatbot (Gemini, Claude, or ChatGPT) to help you think — check a rule, test an idea — but submitting AI-generated answers as your own is not allowed; if AI helped you think, add a one-line note of which tool and how. (Note: this is the traditional format. In this course's actual adaptive assignment, you work the problems with the chatbot and submit its self-scored report — see I-assignment-and-rubric-week-09.md.)

Rubric — 100 points

Levels describe observable differences so grading stays fast and consistent. (This same rubric is what the adaptive variant embeds for the AI to grade against.)

Instructor answer key — REMOVE BEFORE PUBLISHING TO STUDENTS

(All values pre-computed and independently re-verified — w09_verify.py, PASS — 42 checks, 0 failures.)

• Part 1:
  (a) f(x) = 2(x−1)²−3: a=2, h=1, k=−3. Vertex (1, −3), axis x=1, opens up.
  (b) g(x) = x²−4x+7: a=1, b=−4. x=−(−4)/(2)=2; y=4−8+7=3. Vertex (2, 3), axis x=2, opens up.
  (c) h(x) = −3(x+2)²+1: a=−3, h=−2, k=1. Vertex (−2, 1), axis x=−2, opens down.
  (d) p(x) = 2x²+8x−3: a=2, b=8. x=−8/(4)=−2; y=2(4)+8(−2)−3=8−16−3=−11. Vertex (−2, −11), axis x=−2, opens up.

• Part 2: f(x) = 2x²−4x−1 (= 2(x−1)²−3 expanded).
  (a) f(0) = 0−0−1 = −1 → y-intercept (0, −1).
  (b) Quadratic formula: x = (4 ± √(16+8))/4 = (4 ± √24)/4 = (4 ± 2√6)/4 = (2 ± √6)/2. Exact: x = (2+√6)/2 ≈ 2.22 and x = (2−√6)/2 ≈ −0.22. (Equivalently from vertex form: 1 ± √6/2.)
  (c) Discriminant: (−4)²−4(2)(−1)=16+8=24>0 → two real x-intercepts. ✓

• Part 3:
  (a) f(x)=x²−6x+5: a=1>0 → up → minimum. Vertex y=−4. Minimum = −4.
  (b) g(x)=−2x²+8x+3: a=−2<0 → down → maximum. x=−8/(−4)=2; y=−8+16+3=11. Maximum = 11.
  (c) h(x)=5(x−4)²+2: a=5>0 → up → minimum. Vertex y=2. Minimum = 2.

• Part 4: h(t)=−16t²+64t+80. a=−16, b=64, c=80. a<0 → down → maximum.
  (a) t=−64/(−32)=2 seconds.
  (b) h(2)=−64+128+80=144 feet.
  (c) "The ball reaches its maximum height of 144 feet at t = 2 seconds after launch."

Canvas placement block

canvas_object    = Assignment
title            = "Week 9 Assignment — Reading the Parabola (traditional)"
assignment_group = "Assignments"
points_possible  = 100
grading_type     = points
assignment_type  = traditional
submission_types = [online_upload, online_text_entry]
due_offset_days  = 6
published        = true
rubric_ref       = "week-09-assignment-rubric"
provenance       = "~ Prof. Calloway's edition · Fall 2026 · built with thecoursemaker.com"