import React, { useRef, useState, useEffect } from "react"; import { ArrowDown, Check, BookOpen, TrendingUp, Grid, RefreshCw, } from "lucide-react"; import ParabolaWidget from "../../../components/lessons/ParabolaWidget"; import DiscriminantWidget from "../../../components/lessons/DiscriminantWidget"; import LinearQuadraticSystemWidget from "../../../components/lessons/LinearQuadraticSystemWidget"; import Quiz from "../../../components/lessons/Quiz"; import { QUADRATIC_EQ_QUIZ_DATA } from "../../../utils/constants"; import { Frac } from "../../../components/Math"; interface LessonProps { onFinish?: () => void; } const QuadraticEquationsLesson: React.FC = ({ onFinish }) => { const [activeSection, setActiveSection] = useState(0); const sectionsRef = useRef<(HTMLElement | null)[]>([]); const scrollToSection = (index: number) => { setActiveSection(index); sectionsRef.current[index]?.scrollIntoView({ behavior: "smooth", block: "start", }); }; useEffect(() => { const observer = new IntersectionObserver( (entries) => { entries.forEach((entry) => { if (entry.isIntersecting) { const index = sectionsRef.current.indexOf( entry.target as HTMLElement, ); if (index !== -1) setActiveSection(index); } }); }, { rootMargin: "-20% 0px -60% 0px" }, ); sectionsRef.current.forEach((section) => { if (section) observer.observe(section); }); return () => observer.disconnect(); }, []); const SectionMarker = ({ index, title, icon: Icon, }: { index: number; title: string; icon: any; }) => { const isActive = activeSection === index; const isPast = activeSection > index; return ( ); }; return (
{/* Section 1: Parabolas & Forms */}
{ sectionsRef.current[0] = el; }} className="min-h-screen flex flex-col justify-center mb-24 pt-20 lg:pt-0" >

Parabolas & Quadratic Forms

A quadratic function creates a U-shaped curve called a{" "} parabola. The SAT uses three different but equivalent forms — each form highlights different features. Knowing all three lets you pick the most efficient approach for each question.

{/* Three Forms Card */}

The Three Quadratic Forms

Standard Form

y = ax² + bx + c

  • a > 0: parabola opens upward (U-shape);{" "} a < 0: opens downward (∩-shape)
  • |a| > 1: narrow parabola;{" "} |a| < 1: wide parabola
  • c is the y-intercept (the value of y when x = 0)
  • Vertex x-coordinate: x =
  • Axis of symmetry: x =

Example:

y = 2x² − 8x + 6 → axis of symmetry: x ={" "} = ={" "} 2

Vertex Form

y = a(x − h)² + k

  • Vertex is at (h, k) — read directly from the equation
  • Watch the sign: y = a(x − 3)² + 5 has vertex at (3, 5), not (−3, 5)
  • k is the minimum value (if a > 0) or maximum value (if a < 0) of y
  • Best form to use when a question asks about the vertex, minimum, or maximum

Example:

y = −3(x + 2)² + 7 → vertex at (−2, 7); maximum value is{" "} 7 (since a = −3 < 0)

Factored Form

y = a(x − r₁)(x − r₂)

  • x-intercepts (roots/zeros) are at x = r₁ and x = r₂
  • Set each factor = 0 to find roots: x − r₁ = 0 → x = r₁
  • Axis of symmetry is the midpoint of the roots: x ={" "}
  • Best form to use when a question asks about x-intercepts or roots

Example:

y = 2(x − 1)(x − 5) → roots at x = 1 and x = 5; axis of symmetry: x = = 3

{/* Converting Between Forms */}

Converting Between Forms

Standard → Vertex (Completing the Square)

y = x² − 6x + 5

= (x² − 6x + 9) − 9 + 5

= (x − 3)² − 4

Vertex: (3, −4)

Standard → Factored

y = x² − 6x + 5

Find two numbers: × = 5, + = −6

Those numbers: −1 and −5

= (x − 1)(x − 5)

Roots: x = 1 and x = 5

Common SAT Trap — Vertex Form Sign

y = (x + 4)² − 3 looks like h = 4, but it's actually y = (x − (−4))² − 3, so the vertex is at (−4, −3). Always rewrite (x + h) as (x − (−h)) to read the vertex correctly.

{/* Section 2: Solving & Discriminant */}
{ sectionsRef.current[1] = el; }} className="min-h-screen flex flex-col justify-center mb-24" >

Solving Quadratics & The Discriminant

There are four methods to solve a quadratic equation. Choosing the right method quickly is an important SAT skill. Before diving in, check the Discriminant — it tells you instantly how many real solutions exist.

{/* Discriminant Card */}

The Discriminant

Δ = b² − 4ac

For ax² + bx + c = 0

Discriminant Value Number of Real Solutions What it Means Graphically
Δ > 0 2 distinct real solutions Parabola crosses x-axis at two points
Δ = 0 1 repeated real solution Parabola is tangent to x-axis (vertex on x-axis)
Δ < 0 0 real solutions Parabola does not touch x-axis
{/* Four Solving Methods */}

Four Methods to Solve ax² + bx + c = 0

Method 1: Factoring (fastest when it works)

Find two numbers that multiply to a × c and add to b.

Solve: x² + 5x + 6 = 0

Need two numbers: × = 6, + = 5 → 2 and 3

(x + 2)(x + 3) = 0

x = −2 or x = −3

Method 2: Square Root Method (when b = 0 or vertex form)

Isolate the squared term, then take the square root of both sides. Remember ±.

Solve: 2x² − 18 = 0

2x² = 18 → x² = 9

x = ±√9

x = 3 or x = −3

Method 3: Quadratic Formula (always works)

x = −b ± √(b² − 4ac)} d="2a" />

Solve: 2x² − 3x − 2 = 0 (a=2, b=−3, c=−2)

Δ = (−3)² − 4(2)(−2) = 9 + 16 = 25

x = =

x = 2 or x = −½

Method 4: Completing the Square

Rewrite into vertex form, then solve. Most useful when the SAT asks for vertex form or when coefficients are simple.

Solve: x² + 6x + 5 = 0

x² + 6x = −5

x² + 6x + 9 = −5 + 9 (add ( )² = 9 to both sides)

(x + 3)² = 4

x + 3 = ±2

x = −1 or x = −5

{/* SAT Strategy */}

SAT Strategy: Which Method to Use?

1.

If the equation factors nicely (integer roots likely) →{" "} Factor first. It's fastest.

2.

If the middle term is missing (bx = 0) or it's already in vertex form → Square Root Method.

3.

If you can't factor quickly or need exact answers →{" "} Quadratic Formula.

4.

If the question asks to "rewrite in vertex form" or "find the vertex" → Complete the Square.

{/* Section 3: Systems */}
{ sectionsRef.current[2] = el; }} className="min-h-screen flex flex-col justify-center mb-24" >

Linear-Quadratic Systems

When a line intersects a parabola, the system can have 0, 1, or 2 solutions. The key strategy is to substitute the linear equation into the quadratic, rearrange everything to one side to form a new quadratic, then use the discriminant to determine the number of intersections.

Strategy: 4-Step Process

{[ { step: "1", title: "Write out the system", body: "You have a linear equation (y = mx + b) and a quadratic (y = ax² + bx + c). Make sure both are in y = form.", }, { step: "2", title: "Set the right sides equal", body: "Since both equal y, set them equal to each other: mx + b = ax² + bx + c.", }, { step: "3", title: "Rearrange to zero", body: "Move everything to one side: 0 = ax² + (b−m)x + (c−b). You now have a new quadratic equation.", }, { step: "4", title: "Use the Discriminant on the new quadratic", body: "Δ > 0: line crosses parabola at 2 points. Δ = 0: line is tangent (1 point). Δ < 0: line misses parabola (0 points).", }, ].map((item) => (
{item.step}

{item.title}

{item.body}

))}

Worked Example

Find all intersections of y = 2x + 1 and y = x² − 2x + 3.

Set equal: 2x + 1 = x² − 2x + 3

Rearrange: 0 = x² − 4x + 2

Discriminant: Δ = (−4)² − 4(1)(2) = 16 − 8 = 8

Δ > 0 → 2 intersection points

x = = 2 ± √2

Key Distinction: Tangent vs. Intersects

When the SAT says the line is tangent to the parabola, that means exactly 1 intersection → set Δ = 0 and solve for the unknown constant.

{/* Section 4: Quiz */}
{ sectionsRef.current[3] = el; }} className="min-h-screen flex flex-col justify-center" >

Practice Time

{QUADRATIC_EQ_QUIZ_DATA.map((quiz) => (
))}

Topic Mastered!

); }; export default QuadraticEquationsLesson;