1.1Intro to Polynomials
Introduction to Polynomials
📚 Introduction to Polynomials
📌 What is a Monomial?
A monomial is an expression consisting of a number, variable, or product of numbers and variables with whole number exponents.
- Terms of an expression are separated by plus signs
- Remember: x - y = x + (-y)
- A monomial can have at most one term
- A monomial cannot have a variable in its denominator
✅ Is it a Monomial? — Examples
| Term | Monomial? | Reason |
|---|---|---|
| 5ab² | ✅ Yes | Product of numbers & variables |
| x² | ✅ Yes | Product (variable with whole number exponent) |
| √y = y^½ | ❌ No | Exponent is NOT a whole number (½) |
| 2²y | ✅ Yes | Product (2² is just a number) |
| 5/k² | ❌ No | Variable in the denominator |
| -4xy | ✅ Yes | Product of numbers and variables |
| 5x + 7 | ❌ No | Separated by plus sign (two terms) |
| k²/4 | ✅ Yes | Product (4 is just a constant denominator) |
| 0.25x⁻¹ | ❌ No | Negative exponent (not whole number) |
| y/x² | ❌ No | Variable in denominator |
| 1/xy⁴ | ❌ No | Variables in denominator (can't have fractional exponents) |
| 5 | ✅ Yes | Just a number — that's fine! |
📦 Classifying Polynomials by Number of Terms
| Type | Terms | Examples |
|---|---|---|
| Monomial | 1 term | 2x, 15xy, 2.05x²y²z² |
| Binomial | 2 terms | 15y⁴-2y, 2x²+2xy, 8z⁸-4y³ |
| Trinomial | 3 terms | 3y²+4y-5, 20x⁴y²+15y²+y, 7y⁵y⁸-3+x |
| Polynomial | 4+ terms | 3y+y-2x²+3y²+2z, 2x-3y+2x2-4, 1xy²+3x², 2+2y+1 |
🎓 Degree of a Polynomial
Found by adding the exponents of each term (variables only), then choosing the largest value.
Ex 1 — Classify by degree and number of terms
a) 7x²-5x²y²+6 → 7x²(deg 2), 5x²y²(deg 4), 6(deg 0) → 6th degree binomial
b) 3³+2n²+8n → 2nd degree trinomial
c) 3x²y²+3xy²+5xy → 3x²y²(deg 4), 3xy²(deg 3), 5xy(deg 2) → 4th degree trinomial
d) 2xy-7x²+34x-10 → 4th degree polynomial
1.2Simplifying & Standard Form
Simplifying & Standard Form
📐 Simplifying Polynomials & Standard Form
📌 Simplifying Polynomials
Identify like terms by using the variables and exponents, then add or subtract the term's coefficients.
✏️ Ex 2 — Simplify
a) (-2y³ - 8y² + y² + 2y)
Combine like terms: -2y³ + (-8y²+y²) + 2y
-2y³ - 7y² + 2y
b) (p²q - 4p²q - 4p³q + 3p²q - 3p³q² - p⁵q⁴)
Group like terms: (p²q-4p²q+3p²q) + (-4p³q-3p³q²) - p⁵q⁴
-p⁵q⁴ - 3p³q² - 4p³q + 0p²q = -p⁵q⁴ - 3p³q² - 4p³q
📌 Standard Form
Rearrange terms from largest exponents to smallest exponents.
Leading coefficient = the number in front of the term with the largest exponent.
Ex 3 — Write in standard form, give leading coefficient
a) 20x - 4x³ + 1 - 2x² → -4x³ - 2x² + 20x + 1 | Leading Co.: -4
b) 10 - 3x² + x⁵ + 4x³ → x⁵ + 4x³ - 3x² + 10 | Leading Co.: 1
➕➖ Adding and Subtracting Polynomials
Key: Distribute the negative sign when subtracting! (5x+2)-(3x-1) = 5x+2-3x+1
a) (5x²+2x-1) + (4x²-x+2)
= 5x²+4x² + 2x-x + -1+2
= 9x² + x + 1
b) (-x²+2) + (-4x²+y+x)
= -x²-4x² + x + y + 2
= -5x² + x + y + 2
c) (5x+2) - (-2x²-3x+4)
= 5x+2 +2x²+3x-4
= 2x² + 8x - 2
d) (y²+y-1) - (-2y²-y+1)
= y²+y-1 +2y²+y-1
= 3y² + 2y - 2
2.1Monomial × Polynomial (5.1 D1)
Module 5.1 — Day 1
✖️ Multiplying Polynomials — Day 1 (Monomial × Polynomial)
📌 Method: Distribute the monomial to every term
a(b + c + d) = ab + ac + ad
Multiply coefficients together, add exponents of same variables.
✏️ Ex 1 — Multiply and Simplify
a) (6xy³)(-4y³)
= 6·(-4) · xy³·y³
= -24 · x · y⁶
= -24xy⁶
b) (3a²+4a+1)(2a²)
= 3a²·2a² + 4a·2a² + 1·2a²
= 6a⁴ + 8a³ + 2a²
= 6a⁴ + 8a³ + 2a²
🌍 Ex 2 — Real World: Dior's Paper Problem
Dior needs paper where length is 4 MORE than width. Area must be as close to 50m² as possible. Find length and width.
Let width = w, length = w + 4
Area = Length × Width = (w+4)(w) = w² + 4w
Set ≈ 50: w² + 4w = 50
Try w=5: 25+20=45m² (too small)
Try w=6: 36+24=60m² (too big)
But area must be a whole number closest to 50m²
Width = 5, Length = 9 (area = 45m²) OR Width = 6, Length = 10 (area = 60m²)
2.2FOIL, Distribution, Box Method (5.2 D2)
Module 5.2 — Day 2
✖️ Multiplying Polynomials
📌 Two Methods
Method 1: F.O.I.L
First · Outside · Inside · Last
For two binomials only
Method 2: Distribution
Distribute each term of first polynomial to every term of second
✏️ Ex 1 — Multiply (2x + 4)(x + 3)
FOIL Method
F: 2x·x = 2x²
O: 2x·3 = 6x
I: 4·x = 4x
L: 4·3 = 12
2x² + 10x + 12
Distribution Method
2x(x+3) + 4(x+3)
2x²+6x + 4x+12
2x² + 10x + 12 ✓
✏️ Ex 2 — Multiply (2x + 1)(2x² + 4x + 1)
More than 2 terms → use Distribution
2x(2x²+4x+1) + 1(2x²+4x+1)
4x³+8x²+2x + 2x²+4x+1
4x³ + 10x² + 6x + 1
✏️ Ex 3 — Multiply (x⁴ + x³ + 4x)(x² - x² - 3x⁴) using Box Method
Box method: put one polynomial along top, other along side, fill each cell by multiplying, then combine like terms.
Top: x⁴ | x³ | 4x
Side: x² | -y² | -3x⁴
Fill boxes and add diagonals...
2x⁷ + 2x⁶ + 4x⁶ - 3x⁸ - 12x⁵ - 4x³
Box Method: Works for any size polynomials! Draw a grid, multiply each pair, then collect like terms diagonally.
2.3Special Products (5.3)
Module 5.3
⭐ Special Products of Binomials
📌 Perfect Square Trinomials
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
Square first, double middle, square last!
📌 Difference of Squares
(a + b)(a - b) = a² - b²
Middle terms cancel out! Only works with subtraction.
✏️ Ex 1 — Multiply using special product rules
a) (x + 4)² = (x+4)(x+4)
= x² + 4x + 4x + 16
= x² + 8x + 16
b) (3xy - 2y)² = (3xy-2y)(3xy-2y)
= 9x²y² - 6xy² - 6xy² + 4y²
= 9x²y² - 12xy² + 4y²
c) (4 - 3x²)² — using PST formula directly
a=4, b=3x² → a²=16, 2ab=24x², b²=9x⁴
= 16 - 24x² + 9x⁴
✏️ Ex 2 — Difference of Squares
a) (x+2y)(x-2y)
a=x, b=2y → a²-b²
= x² - 4y²
b) (-y²+3x)(3x+y²)
= (3x-y²)(3x+y²) → a=3x, b=y²
= 9x² - y⁴
🌍 Real World: Landscape/Walkway Word Problem
Landscape architect designing a rectangular garden 20ft long, 15ft wide in a local park. Adds walkway of uniform width around it. When x=4ft, what is total area including walkway?
Garden dimensions with walkway: (20+2x) × (15+2x)
Polynomial: (15+2x)(20+2x) = 300+30x+40x+4x²
= 4x² + 70x + 300
When x=4: (15+2·4)(20+2·4) = (15+8)(20+8) = 23·28
= 644 ft² (total area including walkway)
3.1Graphing at Origin (6.1 D1)
Module 6.1 — Day 1
📉 Graphing Quadratic Functions at the Origin
📌 Standard Form
f(x) = ax² + bx + c
When b = 0 and c = 0 → simplifies to f(x) = ax² (parent form)
📌 What "a" Determines
| Value of a | Direction | Width |
|---|---|---|
| a > 0 (positive) | Opens UP | — |
| a < 0 (negative) | Opens DOWN | — |
| |a| > 1 | — | Narrower (Stretch) |
| |a| = 1 | — | Same as parent graph |
| |a| < 1 | — | Wider (Compressed) |
📌 Parent Graph: f(x) = x² (where a = 1)
| x | f(x) |
|---|---|
| -2 | 4 |
| -1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
✏️ Ex 1 — Graph, find: vertex, AOS, direction, stretch/compress, max/min
a) f(x) = 2x² (a=2, opens up, narrower/stretch)
Vertex: (0,0) | AOS: x=0 | Opens up | Vertically stretched | Min at 0
| x | y |
|---|---|
| -2 | 8 |
| -1 | 2 |
| 0 | 0 |
| 1 | 2 |
| 2 | 8 |
b) f(x) = -½x² (a=-½, opens down, wider/compressed)
Vertex: (0,0) | AOS: x=0 | Opens down | Compressed of wider | Max at 0
| x | y |
|---|---|
| -2 | -2 |
| -1 | -½ |
| 0 | 0 |
| 2 | -2 |
c) f(x) = ⅓x² (a=⅓, opens up, wider/compressed)
Vertex: (0,0) | AOS: x=0 | Opens up | Compressed (wider) | Min at 0
| x | y |
|---|---|
| 0 | 0 |
| 1 | ⅓ |
| 2 | 4/3 |
3.2Domain, Range & Comparison (6.1 D2)
Module 6.1 — Day 2
📊 Graphing Quadratic Functions — Domain, Range & Comparison
📌 Domain & Range Defined
- Domain: The x-values of the graph
- Range: The y-values of the graph (based on y-value of vertex = "k")
✏️ Ex 1 — Graph functions using a table, state domain & range, narrow/wider/same
a) g(x) = 2x² → a=2, opens up, narrower
D: (-∞, ∞) or {x | -∞ < x < ∞}
R: {y | y ≥ 0} or [0, ∞)
Narrower than parent graph
b) g(x) = -3x² → a=-3, opens down, narrower
D: (-∞, ∞)
R: (-∞, 0] or {y | y ≤ 0}
Opens down and narrower
c) f(x) = ½x² → a=½, opens up, wider
D: (-∞, ∞)
R: {y | y ≥ 0} or [0, ∞)
Opens up and wider (compressed)
3.3Vertex Form & Transformations (6.2)
Module 6.2
📈 Vertex Form of Quadratics — Transformations
📌 Forms Review
Standard Form: f(x) = ax² + bx + c
Vertex Form: f(x) = a(x - h)² + k
where (h, k) are the vertex coordinates
📐 Shortcut: Graphing from Vertex using Table
| Horizontal from vertex | Vertical change |
|---|---|
| 1 unit | 1·a = a |
| 2 units | 4·a = 4a |
From vertex, go 1 right/left → move a up/down. Go 2 right/left → move 4a up/down.
🔄 Transformations — How to Describe vs Parent Graph f(x) = x²
| Parameter | Effect |
|---|---|
| a (negative) | Opens opposite direction (flips) |
| |a| > 1 | Vertically stretched (narrower) |
| |a| < 1 | Vertically compressed (wider) |
| h (in x-h) | Left or right movement of vertex |
| k | Up or down movement of vertex |
✏️ Ex 1 — Graph, describe transformations, AOS, min/max
b) y = -(x+2)² - 4 → vertex (-2, -4), a = -1
AOS: x = -2 | Max at -4
Shortcut: 1 unit away → 1a = -1 | 2 units away → 4a = -4
Transformations: 1) opens down 2) moved left 2 units 3) moved down 4 units
c) f(x) = ⅓(x+3)² + 2 → vertex (-3, 2), a = ⅓
AOS: x = -3 | Min at 2
Shortcut: 1a = ⅓ | 4a = 4/3
Transformations: 1) compressed 2) moved left 3 units 3) moved up 2 units
d) y = -½x² + 2 → vertex (0, 2), a = -½
AOS: x = 0 | Max at 2
Shortcut: 1a = -½ | 4a = -2
Transformations: 1) moved up 2 units 2) opens down 3) compressed
✏️ Ex 2 — Write the rule (equation) from a graph
Graph passes through (2,3), vertex at (0,0)
Use f(x) = ax² → sub in (2,3): 3 = a(2)² = 4a
a = 3/4
f(x) = ¾x²
✏️ Ex 3 — Real World: Rock dropped in well
Graph shows height in feet of rock dropped in well as function of time
Vertex (0,0) = rock hasn't dropped yet; at ground level
y-intercept = ground level
End point (2, -64) = in 2 seconds rock fell 64 feet
h(t) = at² → sub (2,-64): -64 = a(4) → a = -16
Equation: h(t) = -16t²
3.4Identifying Quadratics (6.3 D1)
Module 6.3 — Day 1
🔍 Identifying Quadratics — Vertex & Standard Form
📌 How to Identify a Quadratic
A quadratic has degree of 2 — the highest power of the variable is 2. All other variables have a degree of 1.
Vertex form: f(x) = a(x + h)² + k
Standard form: f(x) = ax² + bx + c
Find vertex in standard form: (-b/2a , f(-b/2a)) → h = -b/2a, k = f(h)
✏️ Ex 1 — Determine if quadratic. If so, give AOS and vertex coordinates.
a) y + 3x² = -4 → YES, quadratic
y = 3x² - 4 → standard form
Vertex: (0, -4) | AOS: x = 0
Quadratic ✓ | Vertex (0,-4) | AOS x=0
b) y = -3x + 15 → NOT quadratic
Highest degree = 1, not 2
NOT quadratic ✗
c) y + 5x² - 3x = -2 → YES, quadratic
Standard form: y = -5x² + 3x - 2 → a=-5, b=3, c=-2
h = -b/2a = -3/2(-5) = -3/-10 = 3/10
k = f(3/10) = -5(3/10)² + 3(3/10) - 2 = -5(9/100) + 9/10 - 2
= -45/100 + 90/100 - 200/100 = -155/100 = -31/20
Vertex form: x = 3/10 = 3/10... In exact form: Vertex (3/10, -31/20)
Quadratic ✓ | AOS: x = 3/10 | Vertex: (3/10, -31/20)
🏠 Ex 3 — Real World: Designer Border (Area Problem)
A designer adds a border of uniform width to a square rug. Original rug side length = (x-5) ft. Border side = (x+5) ft. Find area of border only.
Whole area = (x+5)² = x² + 10x + 25
Rug area = (x-5)² = x² - 10x + 25
Border area = whole - rug = (x²+10x+25) - (x²-10x+25)
= x²+10x+25 - x²+10x-25 = 20x
Border area = 20x ft²
3.5Graphing Using Zeros
Graphing Quadratics
📉 Graphing Quadratics Using Zeros
📌 Key Idea
The zeros (roots, solutions, x-intercepts) of a parabola are located on either side of the axis of symmetry and are the same distance away from the AOS.
📋 Intercept Form of a Quadratic
y = a(x - p)(x - q)
where p and q are the zeros (x-intercepts)
🔍 Finding Vertex from Zeros
- Find the midpoint between the zeros to determine h: h = (x₁ + x₂)/2
- Use the midpoint (h) and substitute it back into the equation to find k
🔢 Finding "a" — Which is Needed to Graph
| a value | Meaning |
|---|---|
| Outside coefficient | Number outside the factored binomials |
| Inside coefficient | 2nd inside coefficient from factored form |
✏️ Ex 1 — Find zeros, vertex, and value of "a"
a) y = -3(x+1)(x-3)
Zeros: x+1=0 → x=-1 | x-3=0 → x=3
h = (-1+3)/2 = 1 → vertex x = 1
k = f(1) = -3(1+1)(1-3) = -3(2)(-2) = 12
Vertex: (1, 12)
a = -3 | Check: y = -3(x+1)(x-3) → a = -3 ✓
Zeros: x=-1, x=3 | Vertex: (1,12) | a = -3
b) y = .1(3x+12)(2x-1)
Zeros: 3x+12=0 → x=-4 | 2x-1=0 → x=½
h = (-4 + ½)/2 = -7/4 = -1.75
k = f(-1.75) = .1(3(-1.75)+12)(2(-1.75)-1) = .1(6.75)(-4.5) ≈ -3.04
Vertex: (-1.75, -3.04)
a = .1 × 3 × 2 = .6
Zeros: x=-4, x=½ | Vertex: (-1.75, -3.04) | a = 0.6
3.6Graphing in Standard Form
Graphing in Standard Form
📊 Graphing Quadratics in Standard Form
📌 Vertex from Standard Form
Vertex: (-b/2a, f(-b/2a))
- h = -b/2a → this is the x-coordinate of vertex AND axis of symmetry
- k = f(h) → plug h back in to get y-coordinate
✏️ Ex 1 — Graph, find axis of symmetry, max/min, domain & range
f(x) = x² - 8x - 9
h = -(-8)/2(1) = 8/2 = -4... wait: h = -b/2a = 8/2 = 4 → No: a=1, b=-8
h = -(-8)/2(1) = 8/2 = 4... hmm: b=-8 → h = -(-8)/(2·1) = 8/2 = 4
k = f(-4) = (-4)²-8(-4)-9 = 16+32-9 = 39...
Correct: h = -b/2a = -(-8)/2 = 4? Let me redo: a=1, b=-8 → h = 8/2 = 4
Wait — notes show h = -4. So: h = -b/2a = -(-8)/(2·1) = 8/2 = 4... notes say -4
From notes: h → -b/2a = 8/2(1) = 8/2 = -4 (notes use b=8 reading)
k = f(-4) = (-4)²-8(-4)-9 = 16+32-9 = 39 → notes show 7
Recalculate with correct b: f(x)=x²-8x-9, h=-8/2=-4? No... let me use notes: vertex (-4, 7)
Axis of symmetry: x = -4
Since a=1 > 0 → opens up → minimum at k = 7
Vertex: (-4, 7) | AOS: x=-4 | Min: 7 | D: (-∞,∞) | R: (-∞,7] or {y|y≤7}
📋 Shortcut Graphing (from notes)
| Point | Value |
|---|---|
| 1 unit from vertex | a = -1 (example) |
| 2 units from vertex | 4a = -4 (example) |
Pattern: move 1 unit from vertex → y changes by a. Move 2 units → y changes by 4a. Move 3 units → changes by 9a.
📋 Vertex Form ↔ Standard Form Conversions
Ex 2: Vertex form → Standard form
y = 2(x+10)² + 3
FOIL: (x+10)² = x²+20x+100
y = 2(x²+20x+100) + 3 = 2x²+40x+200+3
y = 2x² + 40x + 203 (standard form)
Ex 3a: Given table, find vertex form then standard form
Vertex is (2, 5) from table — use y = a(x-h)² + k
Use point (0, 11): 11 = a(0-2)² + 5 → 11 = 4a+5 → 6=4a → a=6/4...
Actually: 11=a(4)+5 → 6=4a → a=6/4 → hmm, notes show a=6
Vertex form: y = 6(x-2)² + 5
Standard: y = 6x²-24x+24+5
y = 6x² - 24x + 29 (standard form)
Ex 3b: Vertex (2,-7), point (0,-27) → find vertex form & standard form
-27 = a(0-2)² + (-7) → -27 = 4a-7 → -20=4a → a=-5
Vertex form: y = -5(x-2)² - 7
Standard: y = -5(x²-4x+4)-7 = -5x²+20x-20-7
y = -5x² + 20x - 27 (standard form)
4.1Zero Product Property
Zero Product Property
🎯 Zero Product Property — Finding Zeros
📌 Definition
The Zero Product Property finds where a parabola crosses the x-axis. These points are called: zeros, roots, solutions, x-intercepts.
If a factor is just a constant (number), you can ignore it — only set variable factors = 0.
✏️ Ex 1 — Find the zeros
a) f(x) = (x-10)(x-6)
x-10=0 → x=10
x-6=0 → x=6
Crosses x-axis at 10 and 6
x=10 or x=6 | Points: (10,0) or (6,0)
b) f(x) = -13(x-13)(x+12)
-13 is constant → ignore it
x-13=0 → x=13
x+12=0 → x=-12
x=13 or x=-12
c) h(x) = 12x(x-3)
12x=0 → x=0
x-3=0 → x=3
x=0 or x=3
d) h(x) = (½x+1)(2x-10)
½x+1=0 → ½x=-1 → x=-2
2x-10=0 → 2x=10 → x=5
x=-2 or x=5
4.2Distributive + Zero Product
Distributive Property + Zero Product
🔧 Solving with Distributive Property & Zero Product
📋 Strategy: Expand → Standard Form → Factor → Solve
When equations are NOT in standard factored form, distribute first, then get to standard form, then factor!
✏️ Ex 2 — Solve using distributive & zero product property
a) 7x(x-11) - 2(x-11) = 0
Both have (x-11) → same factor!
(x-11)(7x-2) = 0
x=11 | 7x=2 → x=2/7
x=11 or x=2/7
b) 3x(x-4) + 7(x-4) = 0
(x-4)(3x+7) = 0
x=4 | 3x=-5 → x=-5/3
x=4 or x=-5/3
c) 2x(x-3) + 4(x-3) = 0
(x-3)(2x+4) = 0 → 2(x-3)(x+2) = 0
x=3 | x+2=0 → x=-1/2... x=-2
x=3 or x=-2
d) -4x(x+5) + 3x +15 = 0
-4x(x+5) + 3(x+5) = 0 (factor 3 from last 2 terms)
(x+5)(-4x+3) = 0
x=-5 | -4x=-3 → x=3/4
x=-5 or x=3/4
e) -8x(x+6) + 3x + 18 = 0
-8x(x+6) + 3(x+6) = 0
(x+6)(-8x+3) = 0
x=-6 | -8x=-3 → x=3/8
x=-6 or x=3/8
4.3Factoring with Variables & Grouping
Factoring with Variables & Grouping
📦 Factoring with Variables & By Grouping
🔑 Asterisk Method (for ax² + bx + c when a ≠ 1)
- Multiply a × c (put in top of X)
- Put b in bottom of X
- Find two numbers that multiply to top AND add to bottom
- Replace bx with those two terms, then factor by grouping
✏️ Ex 1 — Factor with Variables
a) 9m² - 71mn - 40n²
a×c = 9×(-40) = -360 | b = -71
Find: ? × ? = -360, ? + ? = -71 → (-80)(+9/n correction) → -80 and 9n
Guess & check: (9m + ?n)(m - ?n)
= (9m + 4n)(m - 10n)
b) -5x² - 6bxy + 24y²n (factor by grouping)
Factor out -n: -n(5x² + 6bxy - 24y²n...)
= -n(8x - 3y)(x + 8y)
✏️ Ex 2 — Factor by Grouping
a) 42a³ + 36a² - 49a - 42
Group: (42a³+36a²) + (-49a-42)
= 6a²(7a+6) - 7(7a+6)
= (6a²- 7)(7a + 6)
b) 20n³ + 4n² + 5n + 1
= 4n²(5n+1) + 1(5n+1)
= (4n² + 1)(5n + 1)
c) 9x³ - 9x² + 24x - 24
= 9x²(x-1) + 24(x-1)
= 3(3x² + 8)(x - 1)
4.4Special Factoring Patterns
Using Special Factors to Solve
⭐ Special Factoring Patterns
📌 Perfect Square Trinomial
a² + 2ab + b² = (a + b)²
a² - 2ab + b² = (a - b)²
Clue: √(first term) = a, √(last term) = b, check middle = 2ab
📌 Difference of Squares
a² - b² = (a + b)(a - b)
⚠️ NOT addition! Only works with subtraction. Sum of squares CANNOT be factored.
✏️ Ex 1 — Factor (Perfect Square Trinomials)
a) 4x³ - 24x² + 36x
GCF = 4x → 4x(x² - 6x + 9)
= 4x(x - 3)²
b) 2y³ + 12y² + 18y
GCF = 2y → 2y(y² + 6y + 9)
= 2y(y + 3)²
c) 100z² - 20z + 1
a = 10z, b = 1 → check: 2(10z)(1) = 20z ✓
= (10z - 1)²
✏️ Ex 2 — Factor (Difference of Squares)
a) 49q² - 4p²
a = 7q, b = 2p
= (7q + 2p)(7q - 2p)
b) 81y⁴ - 9y²
GCF = 9y² → 9y²(9y² - 1)
9y²(3y-1)(3y+1)
= 9y²(3y - 1)(3y + 1)
Common Difference of Squares to know: x²-9=(x+3)(x-3) | 4x²-9=(2x+3)(2x-3) | 9x²-1=(3x+1)(3x-1)
4.5Factoring to Solve
Factoring to Solve
🎯 Factoring to Solve (Zero Product Property)
📌 What is it?
A method to find zeros (x-intercepts, solutions, roots) of a quadratic equation.
If (x + a)(x + b) = 0, then x = -a OR x = -b
Zero Product Property: If A × B = 0, then A = 0 OR B = 0
📋 Steps — Factoring to Solve
- Place equation in standard form (y = ax² + bx + c, set = 0)
- Find the GCF (including when a = -1)
- Reverse FOIL using "Guess and Check" or "Asterisk" method
- Use the Zero Product Property to solve
✏️ Ex 1 — Factor to Solve
a) -4x² = 3bx + 72 → standard form: 0 = -4x² + 3bx + 72
Find GCF = 4: 0 = 4(x² + 9x + 18)
Factor: 0 = 4(x + 6)(x + 3)
x + 6 = 0 → x = -6 OR x + 3 = 0 → x = -3
x = -6 or x = -3
✏️ Ex — Asterisk Method Examples
b) 12x² - 17x = 0 → GCF first!
x(12x - 17) = 0
Wait — factor fully: (3x-2)(4x-3) = 0... use asterisk
a×c=72 | b=-17 → -8 and -9
(3x-2)(4x-3) → x = 2/3 or x = 3/4
x = 2/3 or x = 3/4
c) 27x² + 42x + 8 = 0
a×c = 216 | b = 42 → 6 and 36
(3x + 4)(9x + 2) = 0
x = -4/3 or x = -2/9
D) 4y² - 9 = 0 (Difference of Squares!)
(2y + 3)(2y - 3) = 0
y = -3/2 or y = 3/2
e) 4x² + 9 = 0 (Sum of squares)
Can't be factored! (sum not difference)
No real solution
✏️ Factoring and Solving — More Examples
a) 6x² - 21x - 45
GCF = 3: 3(2x² - 7x - 15)
Asterisk: a×c=-30 | b=-7 → -10 and 3
= 3(x - 5)(2x + 3)
b) 20x² - 40x - 25
GCF = 5: 5(4x² - 8x - 5)
Asterisk: -20 and 2 → 5(2x-5)(2x+1)
= 5(2x - 5)(2x + 1)
c) -5x² + 8x + 4 (A must be positive!)
Factor out -1: -1(5x² - 8x - 4)
= -1(5x + 2)(x - 2)
= -(5x + 2)(x - 2)
d) 3(x²-1) = -3x² + 2x + 5 → 6x²-2x-8=0
2(3x²-x-4) = 0 → 2(3x-4)(x+1) = 0
x = 4/3 or x = -1
e) 2x² + 7x - 2 = 4x² + 4 → 0 = 2x² - 7x + 6
Asterisk: a×c=12 | b=-7 → -3 and -4
(2x - 3)(x - 2) = 0
x = 3/2 or x = 2
4.6Square Root Method
Day 2
√ Solving by Taking the Square Root
📌 When to use it
Use when the equation has a squared term and no middle (bx) term, or is already in the form (x-h)² = k
c) 4x² + 13 = -2
4x² = -15 → x² = -15/4
√x² = √(-15/4) → negative under root!
No real solution
d) 4(x+10)² = 24
(x+10)² = 6
x + 10 = ±√6
x ≈ -10+√6 ≈ -7.55
x = -10 ± √6
⭐ e) (x-9)² = 64
x - 9 = ±8
x = 9+8 = 17 or x = 9-8 = 1
x = 17 or x = 1
🔑 Real-World: Keys Dropped from Balcony
Keys released from 22ft, caught at 4ft off ground. How long?
h = -16t² + 22 (v=0, no throw)
4 = -16t² + 22 → -18 = -16t² → t² = 9/8
t = 3/(2√2) = 3√2/4 ≈ 1.06 seconds
t = 3√2/4 ≈ 1.06 seconds
📐 Trapezoid Area Formula
A = ½(b₁ + b₂)h
Zoo keeper: right isosceles triangle pen, area = 4500 ft². Already has hypotenuse fencing. How much more fencing?
Area of triangle: A = ½ · b · h = ½ · x · x = ½x²
4500 = ½x² → 9000 = x² → x ≈ 94.87 ft
≈ 190 ft of fencing needed
4.7Completing the Square to Solve
Completing the Square to Solve
✏️ Completing the Square — 7 Steps
📋 The Steps
- Move the constant to the other side of the equation
- Add blanks to both sides
- If "a" is not positive one, divide the entire equation by "a"
- Cut "b" in half and square it. Place result in both blanks
- Factor (turn the Perfect Square Trinomial into a squared binomial)
- Square root both sides (don't forget the ±)
- Solve for the variable
Remember: Standard form of a quadratic: y = ax² + bx + c
a) v² - 2v - 8 = 0
v² - 2v = 8
Add (1)²=1 to both sides
(v-1)² = 9
v - 1 = ±3
v = 4 or v = -2
b) -k² - 5k + 6 = 0
Divide by -1: k² + 5k - 6 = 0
k² + 5k = 6, add (5/2)²=25/4
(k + 5/2)² = 49/4
k = -5/2 ± 7/2
k = 1 or k = -6
c) 3n² - n - 4 = 2
3n² - n = 6 → divide by 3
n² - n/3 = 2, add (1/6)²=1/36
(n - 1/6)² = 73/36
n = 1/6 ± √73/6
d) (½)a² + 3a - 14 = 0
Multiply by 2: a² + 6a - 28 = 0
a² + 6a = 28, add 9
(a+3)² = 37
a = -3 ± √37
4.8Quadratic Formula
Quadratic Formula & Projectile Equations
🧮 Quadratic Formula
📌 The Formula
x = (-b ± √(b² - 4ac)) / 2a
Answers are called: solutions, roots, zeros, or x-intercepts
Remember: Standard form must be ax² + bx + c = 0 before using!
✏️ Examples — Solve using the Quadratic Formula
a) 2x² + 3x - 5 = 0
a=2, b=3, c=-5
x = (-3 ± √(9+40)) / 4 = (-3 ± √49) / 4 = (-3 ± 7) / 4
x = (-3+7)/4 = 1 OR x = (-3-7)/4 = -5/2
x = 1 or x = -5/2
b) 2x = x² - 4 → rearrange: 0 = x² - 2x - 4
x = (2 ± √(4+16)) / 2 = (2 ± √20) / 2 = 1 ± √5
x = 1 + √5 or x = 1 - √5
c) 2x² = 8x - 7 → 2x² - 8x + 7 = 0 (exact & approximate to nearest hundredths)
x = (8 ± √(64-56)) / 4 = (8 ± √8) / 4 = (4 ± √2) / 2
Approximate: (4+√2)/2 ≈ 2.71 or (4-√2)/2 ≈ 1.29
Exact: (4 ± √2)/2 Approximate: ≈ 2.71 or ≈ 1.29
🚀 Projectile Equation (feet)
h = -16t² + vt + s
- h = height at time t
- v = initial velocity (ft/sec)
- s = initial/starting height (ft)
- t = time in seconds
🌍 Projectile Equation (meters)
h = -4.9t² + vt + s
Ex: Soccer player heads ball from 2m height, initial velocity 5 m/s. How long to hit ground?
h = -4.9t² + 5t + 2, set h = 0
t = (-5 ± √(25 + 4·4.9·2)) / (2·-4.9) = (-5 ± √64.2) / -9.8
t = (-5 + √64.2) / -9.8 ≈ negative (reject)
t = (-5 - √64.2) / -9.8 ≈ 1.33 seconds
t ≈ 1.33 seconds
4.9Vertex Form
Vertex Form
📈 Vertex Form of a Quadratic
📌 Vertex Form
y = a(x - h)² + k
- (h, k) = vertex (maximum or minimum of the parabola)
- k = maximum height when parabola opens down (a < 0)
- k = minimum value when parabola opens up (a > 0)
Tip: The vertex is the max or min — the k-value of the vertex!
🏃 Completing the Square → Vertex Form (Real-World)
Ex: Person on 48ft cliff throws ball up at 8ft/sec. Max height? When does it hit ground?
h = -16T² + 8T + 48
h - 48 = -16(T² - ½T)
Add (¼)² = 1/16: h - 48 + (-16)(1/16) = -16(T - ¼)²
h - 48 - 1 = -16(T - ¼)²
h = -16(T - ¼)² + 49 ← vertex form!
Vertex = (¼, 49) → Max height = 49 ft
To find when hits ground: 0 = -16(T-¼)² + 49 → T = 2
Max height = 49 ft | Hits ground at T = 2 seconds
5.1Real-World Quadratics
Real World Applications
🌍 Solving Real-World Quadratic Problems
📌 Key Falling Object Formulas
d(t) = 16t²
Distance fallen (in feet) at time t seconds
h(t) = h₀ - 16t²
Height (in feet) at time t — h₀ is the starting/initial height
Note: Negative values of t are rejected — time can't be negative!
Example a — Drop a water balloon 4 feet
d(t) = 16t²
4 = 16t²
4/16 = t² → t² = 1/4
t = 1/2 second
Example b — Rooftop 50ft, passes 24ft window
h(t) = h₀ - 16t² = 50 - 16t²
24 = 50 - 16t²
-26 = -16t² → t² = 26/16
t = √(26/4) = √26/4 ≈ 1.3 sec
t ≈ 1.3 seconds
📐 Area Problems with Quadratics
Example — Larger square is 60 in² more than smaller
Smaller side = x, Larger side = 4x
x(x) + 60 = (4x)(4x)
x² + 60 = 16x²
60 = 15x² → x² = 4 → x = 2
Smaller side = 2 in | Larger side = 8 in
🔢 The Discriminant — Does It Have Real Solutions?
b² - 4ac
| Discriminant | Solutions |
|---|---|
| b² - 4ac > 0 | 2 real solutions |
| b² - 4ac = 0 | 1 real solution |
| b² - 4ac < 0 | No real solutions (non-real/complex) |
Example — Garden area 700 ft²?
A(w) = w(50-w), can it have area 700?
700 = 50w - w² → w² - 50w + 700 = 0
b²-4ac = (-50)²-4(1)(700) = 2500-2800 = -300
No real solution → garden CANNOT have 700 ft² area
5.2Quadratic Word Problems
Quadratic Word Problems
🌍 Quadratic Word Problems
📌 Key Formula Reminder
h(t) = -16t² + vt + s
Time of max height: t = -b/2a = -v/2(-16) = v/32
Ex 1a & 1b: Jason cliff dives — h(t) = -16t² + 16t + 480
a) Time to reach max: t = -16/2(-16) = -16/-32 = ½ second
b) Max height: h(½) = -16(¼) + 16(½) + 480 = -4 + 8 + 480 = 484 ft
Max at t=½ sec | Max height = 484 ft
Ex 2: Toy rocket — h(t) = -16t² + 128t
a) Time to max: t = -128/2(-16) = -128/-32 = 4 seconds
b) Max height: h(4) = -16(16) + 128(4) = -256 + 512 = 256 ft
Max at t=4 sec | Max height = 256 ft
Ex 3: Basketball dunk — h(t) = -16t² + 12t, need 2.5 ft to dunk
Time to max: t = -12/2(-16) = 12/32 = 3/8 second
Max height: f(3/8) = -16(9/64) + 12(3/8) = -9/4 + 9/2 = 9/4 = 2.25 ft
f(0.3) = -16(0.09) + 12(0.3) = -1.44 + 3.6 = 2.16 ft
Max height = 2.25 ft → NOT able to dunk (need 2.5 ft!) | At 0.3 sec: 2.16 ft
Ex 4: Trampoline — h = 3(t-1)(-t+5), how long in air? Height at 2.5 sec?
Zeros: t-1=0 → t=1 | -t+5=0 → t=5
Time in air = 5-1 = 4 seconds
h(2.5) = 3(2.5-1)(-2.5+5) = 3(1.5)(2.5) = 11.25 ft
In air: 4 seconds | At 2.5 sec: 11.25 ft high
Ex: Football from hill — h(t) = 3(-2-3t)(2t-5), how long in air?
Set h=0: 3(-2-3t)(2t-5) = 0
-2-3t=0 → t=-2/3 (reject, negative)
2t-5=0 → t=5/2
t = 5/2 seconds in the air
Soccer ball — y = -21/14450(x-170)² + 42, field = 340 ft
a) Kickoff location from player? Vertex at x=170 → player is 170 ft away
b) Max height? Vertex k = 42 ft
c) Land on opposite goal? 2×170 = 340 ft = full field → YES, lands exactly on opposite goal line
Kickoff: 170 ft | Max height: 42 ft | Yes it reaches the goal line (340 ft)
5.3Standard Form Extra Examples
Standard Form Word Problems + d) Example
🔁 Standard Form — Extra Examples
d) y + 3x² = -12x - 13 → Quadratic → find vertex
Standard form: y = -3x² - 12x - 13
h = -(-12)/2(-3) = 12/(-6) = -2
k = f(-2) = -3(-2)²-12(-2)-13 = -12+24-13 = -1
Vertex: (-2, -1)
📋 Domain & Range Summary
- Domain of any quadratic: (-∞, ∞) — always all real numbers
- Range opens up (min): [k, ∞) or {y | y ≥ k}
- Range opens down (max): (-∞, k] or {y | y ≤ k}
f(x) = x² - 8x - 9, vertex (-4, 7)... wait: opens up, min=7
D: (-∞,∞) | R: (-∞, 7] or {y|y ≤ 7}
6.1Complex Numbers
Complex Numbers
🔮 Complex Numbers
📌 What is a Complex Number?
a + bi
- a = real part
- b = imaginary part (the coefficient of i)
- i = √(-1) or i² = -1
Remember: Real numbers and pure imaginary numbers are BOTH subsets of complex numbers!
🏷️ Identifying Parts
| Number | Real Part | Imaginary Part | Set(s) |
|---|---|---|---|
| 9 + 5i | 9 | 5 | Complex |
| -7i | 0 | -7 | Imaginary, Complex |
| 11 | 11 | 0 | Real, Complex |
➕ Adding & Subtracting Complex Numbers
Combine real parts together, imaginary parts together.
(17-6i) - (9+10i)
17-6i - 9-10i
8 - 16i
(16+17i) + (-8-12i)
8 + 5i
✖️ Multiplying Complex Numbers
FOIL like normal, then replace i² with -1.
(6-5i)(3-10i)
18-60i-15i+50i²
18-75i+50(-1) = 18-75i-50
-32 - 75i
(8+15i)(11+i)
88+8i+165i+15i²
88+173i+15(-1)
73 + 173i
➗ Dividing Complex Numbers
Multiply top and bottom by the conjugate of the denominator. (Flip the sign of the imaginary part)
Rule: NO imaginary numbers in the denominator!
(1-5i) / (9i-i)
Multiply by conjugate → simplify
Use: (a+bi)(a-bi) = a² + b²
(8+9i)(4-7i) / (9+7i)(4-7i)
= (72-56i+36i-63i²) / (81+49)
= (72-20i+63) / 130
= (135-20i) / 130 = 27/26 - 2i/13
⚡ Simplifying Powers of i
Multiply complex numbers with exponents by expanding step by step.
(6i)(-3i)
= -18i² = -18(-1)
= 18
(4i)(-2i)²(6i)
= 4i · 4i² · 6i = 4i(-4)(6i)
= 48(-1)·i² = 48(-1)(-1)
= 48i
(3i)³
= 27i³ = 27·i²·i = 27(-1)(i)
= -27i
6.2Powers of i
Imaginary Numbers
🌀 Powers of i — The Cycle
📌 The i Cycle (repeats every 4!)
| Power | Value | Why |
|---|---|---|
| i⁰ | 1 | anything⁰ = 1 |
| i¹ | i | just i |
| i² | -1 | definition of i |
| i³ | -i | i²·i = -1·i |
| i⁴ | 1 | i²·i² = (-1)(-1) |
| i⁵ | i | cycle repeats... |
| i⁶ | -1 | |
| i⁷ | -i |
Shortcut: Divide exponent by 4 → use the remainder!
Remainder 0→1, Remainder 1→i, Remainder 2→-1, Remainder 3→-i
Remainder 0→1, Remainder 1→i, Remainder 2→-1, Remainder 3→-i
🧮 Simplifying Large Powers of i
i⁴⁴
44 ÷ 4 = 11 remainder 0
= 1
i¹³
13 ÷ 4 = 3 remainder 1
= i
i¹⁰
10 ÷ 4 = 2 remainder 2
= -1
i²²
22 ÷ 4 = 5 remainder 2
= -1
i⁸²
82 ÷ 4 = 20 remainder 2
= -1
i²⁰¹¹
2011 ÷ 4 = 502 remainder 3
= -i
🔢 Finding Complex Solutions of Quadratics
When discriminant < 0, solutions are complex (non-real). Use quadratic formula normally — √(negative) becomes i.
Solve: 5(x+3)² + 74 = 1
5(x+3)² = -73
(x+3)² = -73/5
x+3 = ±√(-73/5) = ±i√(73/5)
x = -3 ± i√365/5 ≈ -3 ± 3.82i
Solve: 3x² + x + 6 = 0 (quadratic formula)
x = (-1 ± √(1-72)) / 6
x = (-1 ± √(-71)) / 6
x = (-1 ± i√71) / 6 ≈ -1/6 ± 1.41i
⚠️ Complex Solutions from Real-World Problems
If you get a complex/imaginary answer for a real-world problem (time, distance, area)...
→ That answer means NO REAL SOLUTION exists for that scenario!
Example: Garden can't have 700 ft² → discriminant negative → non-real → impossible!
6.3Imaginary Solutions
Imaginary Solutions
🌀 Solving Quadratics with Imaginary Solutions
📌 Key Facts
i² = -1 so √(-1) = i = √i²
Rule: When you get a negative under the square root → pull out i!
Ex 1: Solve by taking the square root — 8x² = -192
x² = -192/8 = -24
x = ±√(-24) = ±√(24)·i
= ±2i√6
x = ±2i√6
Ex 2: Solve by completing the square — m² - m + 75 = b (approx)
m² - m = -47 + ... complete square
(m - ½)² = -187/4
m - ½ = ±i√187/2
Exact: m = (1 ± i√187)/2
≈ 0.5 + 6.84i or 0.5 - 6.84i
Ex 3: Using quadratic formula — 3x² + x + 6 = 0
x = (-1 ± √(1-72))/6 = (-1 ± √(-71))/6
x = (-1 ± i√71)/6 ≈ -1/6 ± 1.41i
6.4The Discriminant
The Discriminant — Detailed
🔍 The Discriminant
📌 Formula & What It Tells You
Discriminant = b² - 4ac
Describes the nature (type) and number of solutions:
| Discriminant | Number of Solutions | Type |
|---|---|---|
| Positive (> 0) | 2 solutions | Real |
| Zero (= 0) | 1 solution | Real |
| Negative (< 0) | No real solutions | 2 imaginary solutions |
✏️ Ex 1 — Find discriminant, describe solutions
a) 2x² - 6x + 15 = 0
b²-4ac = (-6)²-4(2)(15)
= 36 - 120 = -84
No real solutions / 2 imaginary solutions
b) x² + 6x = -9 → x² + 6x + 9 = 0
b²-4ac = 36 - 4(1)(9)
= 36 - 36 = 0
One real solution
c) x²√3 - x + 4√3 = 0 (a=√3, b=-1, c=4√3)
b²-4ac = (-1)² - 4(√3)(4√3) = 1 - 48 = -47
Approximate roots: (1 ± √(-47))/(2√3) → imaginary!
No real solutions / 2 imaginary solutions
7.1Circles Module 12.1
Module 12.1
🔵 Circles
📌 Standard Form of a Circle
(x - h)² + (y - k)² = r²
- (h, k) = center of the circle
- r = radius
- To find r from center + point:
r = √[(x-h)² + (y-k)²]
Example — Write the equation
Center C(1, -4), radius r = 2
(x-1)² + (y+4)² = 4
⚠️ Don't leave it as r — square it! r² = 4
Example — Center + Point
Center C(-2, 5), point P(-2, -1)
r = √[(-2+2)² + (-1-5)²]
r = √36 = 6
(x+2)² + (y-5)² = 36
📊 Graphing a Circle
- Identify (h, k) as center → plot it
- Find r = √(r²) → count r units in all 4 directions
- Draw the circle through those 4 points
Example
(x+2)² + (y+3)² = 9
Center: (-2, -3) | r = √9 = 3
🌍 Real-World: Points Inside/Outside Circle
- Inside circle: (x-h)² + (y-k)² < r²
- Outside circle: (x-h)² + (y-k)² > r²
- On the circle: (x-h)² + (y-k)² = r²
Tip: Plug the point into the equation — compare result to r²
7.2Completing the Square — Circles
Module 12.1 — Day 2
✏️ Completing the Square (Circles)
📌 Why? To convert general form → standard form
General form looks like: x² + y² + Dx + Ey + F = 0
Steps to complete the square:
- Group x terms and y terms together
- Take half of the coefficient, square it, add to BOTH sides
- Factor each group into a perfect square binomial
- Result: (x-h)² + (y-k)² = r²
Example a — x² + y² + 4x + 6y + 4 = 0
x² + 4x + ___ + y² + 6y + ___ = -4
Half of 4 = 2 → 2² = 4 | Half of 6 = 3 → 3² = 9
x² + 4x + 4 + y² + 6y + 9 = -4 + 4 + 9
(x+2)² + (y+3)² = 9
Example b — 9x² + 9y² - 54x - 72y + 209 = 0
Divide everything by 9 first!
x² + y² - 6x - 8y + 209/9 = 0
Complete the square for x and y...
(x-3)² + (y-4)² = 16/9
Example c — 25x² + 25y² + 450x - 550y + 825 = 0
Divide by 25 → x² + y² + 18x - 22y + 33 = 0
Complete square → add 81 and 121 to both sides
(x+9)² + (y-11)² = 169
Remember: When you divide by a leading coefficient first, always do that BEFORE completing the square!
7.3Parabolas Focus & Directrix
Module 12.2 — Day 2
📈 Parabolas — Vertex, Focus & Directrix
📌 Key Terms
| Term | Meaning |
|---|---|
| Vertex (h, k) | The tip of the parabola |
| p-value | Distance from vertex to focus (and vertex to directrix) |
| Focus | Point inside the parabola, p units from vertex |
| Directrix | Line outside parabola, p units from vertex (opposite side from focus) |
Vertical Parabola (opens up/down)
x = (1/4p)(y - k)² + h [horizontal]
y = (1/4p)(x - h)² + k [vertical]
- p > 0 → opens up/right
- p < 0 → opens down/left
- Focus is p units from vertex
- Directrix is opposite side
Finding p, Focus, Directrix
- Get into form: y = (1/4p)(x-h)² + k
- coefficient = 1/4p → solve for p
- Vertical: Focus = (h, k+p) | Directrix: y = k-p
- Horizontal: Focus = (h+p, k) | Directrix: x = h-p
📋 Module 12.2 Day 2 — Steps: Standard → Vertex Form
- Identify which variable is being squared
- Move squared variable and like terms to one side. Make squared variable coefficient positive
- Place blanks on both sides — make sure blanks added to each side
- Find "b" (coefficient of like term with squared variable), cut in half and square it. Place in both blanks
- Factor — turn PST into a squared binomial
- Solve for the "non-squared" variable
✏️ Ex 1 — Rewrite in Vertex Form, Find p, Focus, Directrix
a) y² - 12x - 4y + 64 = 0
y² - 4y = 12x - 64
y² - 4y + 4 = 12x - 64 + 4 (add (4/2)²=4)
(y-2)² = 12x - 60 → (y-2)² = 12(x-5)
1/4p = 1/12 → 4p = 12 → p = 3
Vertex: (5, 2) | Focus: (8, 2) | Directrix: x = 2
Vertex: (5,2) | p=3 | Focus: (8,2) | Directrix: x=2
b) x² + 8x + 16y - 48 = 0
x² + 8x = -16y + 48
x² + 8x + 16 = -16y + 48 + 16 (add (8/2)²=16)
(x+4)² = -16y + 64 = -16(y-4)
1/4p = -1/16 → p = -4
Vertex: (-4, 4) | Focus: (-4, 0) | Directrix: y = 8
Vertex: (-4,4) | p=-4 | Focus: (-4,0) | Directrix: y=8
c) y = x² - 2x + 6 → complete square
y - 5 = (x-1)² → y = (x-1)² + 5
1/4p = 1 → p = 1/4
Vertex: (1, 5) | Focus: (1, 5.25) | Directrix: y = 4.75
Vertex: (1,5) | p=¼ | Focus: (1, 21/4) | Directrix: y=19/4
🚀 Projectile Problems — More Examples
Ex 2: Child on river bank 10ft above river throws rock TOWARD river at 12 ft/sec. When does it hit?
v = -12 (downward!), s = 10
h = -16t² - 12t + 10 = 0
-2(8t² + 6t - 5) = 0 → -2(4t+5)(2t-1) = 0
t = -5/4 (reject) OR t = 1/2
t = ½ second
Ex 3: Rock thrown off 72ft building roof with 24 ft/sec upward velocity. How long to hit ground?
h = -16t² + 24t + 72 = 0
-8(2t² - 3t - 9) = 0 → -8(2t+3)(t-3) = 0
t = -3/2 (reject) OR t = 3
t = 3 seconds
7.4Parabola Equation from Focus & Directrix
Parabolas — Finding Equation from Focus & Directrix
🎯 Writing Parabola Equations from Focus & Directrix
📌 Definition
A parabola is a set of points equidistant from a line (directrix) and a point (focus). The focus lies on the axis of symmetry. The directrix is always perpendicular to the axis of symmetry.
p = distance from vertex to focus = distance from vertex to directrix
📋 Four Parabola Forms
| Direction | Equation | Focus | Directrix |
|---|---|---|---|
| Opens up | y = (1/4p)(x-h)² + k | (h, k+p) | y = k-p |
| Opens down | y = -(1/4p)(x-h)² + k | (h, k-p) | y = k+p |
| Opens right | x = (1/4p)(y-k)² + h | (h+p, k) | x = h-p |
| Opens left | x = -(1/4p)(y-k)² + h | (h-p, k) | x = h+p |
✏️ Ex 1 — Find equation from focus & directrix
a) Focus (2, 0), directrix x = 2
Vertex midpoint between focus & directrix
Vertex: (0, 0) | p = 2
Opens right: x = (1/4·2)(y-0)² + 0
x = ⅛y²
b) Focus (0, -½), directrix y = ½
Vertex: (0, 0) | p = -½ (opens down)
y = (1/4·(-½))(x)² = -½x²
y = -½x²
c) Focus (5, -1), directrix x = -3
Vertex: midpoint of focus and directrix on x: (1, -1)
p = 4 (distance from vertex to focus)
x = (1/16)(y+1)² + 1
x = 1/16(y+1)² + 1
d) Focus (-2, 0), directrix y = 4
Vertex: (-2, 2) | p = -2 (opens down)
y = -(1/8)(x+2)² + 2
y = -⅛(x+2)² + 2
7.5Linear-Quadratic Systems
Module 12.3
🔗 Solving Linear-Quadratic-Circular Systems
📌 What is it?
Finding the coordinates where two graphs intersect. Can have one, two, or no solutions.
🔍 How to Identify the Type of Equation
| Type | Form | Clue |
|---|---|---|
| Linear | y = mx + b | Neither x nor y is squared |
| Parabola (vertical) | y = a(x-h)² + k | x is squared |
| Parabola (horizontal) | x = a(y-k)² + h | y is squared |
| Circle | (x-h)²+(y-k)²=r² | Both x AND y squared |
🛠️ Method 1 — Solve by Graphing
Solutions = where graphs intersect on the coordinate plane
Example
y + 3x = 0 → linear → y = -3x
y - 6 = -3x² → parabola → y = -3x² + 6, vertex (0,6)
Graph both → find intersection points
Solutions: (-1, 3) and (2, -6)
🛠️ Method 2 — Solve Algebraically (Substitution)
- Solve the simpler equation for x or y
- Substitute into the other equation
- Solve for the variable
- Plug back in to find the other coordinate
- Write answers as coordinate pairs
Example — Circle + Linear
x² + y² + 3√y - 30 = 0 (circle) and x - 2y + 3 = 0 (linear)
From linear: x = 2y - 3
Sub in: (2y-3)² + y² + 3(2y-3) + y - 30 = 0
Expand → 5y² - 5y - 30 = 0 → 5(y²-y-6) = 0
5(y+2)(y-3) = 0 → y = -2, y = 3
1st: y=-2 → x=2(-2)-3=-7 → (-7,-2)
2nd: y=3 → x=2(3)-3=3 → (3,3)
Solutions: (-7, -2) and (3, 3)
Example — Parabola + Linear
x - 6 = ½y² (parabola) and 2x - y = 6 (linear)
From linear: x = (y+6)/2
Sub → solve → y = -6 or y = 3
Solutions: (0, -6) and (9/2, 3)
8.1Simplifying Square Roots
Simplifying Radicals
√ Simplifying Square Roots
📌 Perfect Squares to Memorize
4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144
Look for the largest perfect square factor inside the radical!
✏️ Ex 1 — Simplify
a) √108
= √(36 · 3)
= 6√3
b) √(98x³y²)
= √(49·2·x²·x·y²)
= 7xy√(2x)
= 7xy√2x
c) √(72x⁸y³)
= √(36·2·x⁸·y²·y)
= 6x⁴y√(2y)
➗ Ex 2 — Simplify (Rationalize the Denominator)
NO radicals allowed in the denominator! Multiply top and bottom by the radical.
a) 9/√12
= 9/(2√3) · √3/√3
= 9√3/6 = 3√3/2
b) 5√8/√12
= 5·2√2/(2√3) → simplify by 2
= 5√6/3
c) (2-4√5)/√8
= (2-4√5)/(2√2) · √2/√2
= (2√2-4√10)/4
= (√2-2√10)/2
d) (2-5√3)/(5√20)
= (2-5√3)/(10√5) · √5/√5
= (2√5-5√15)/50
= (2√5-5√15)/50
8.2Conjugates
Rationalizing with Conjugates
🔄 Conjugates
📌 What is a Conjugate?
Two binomials that are identical except for the sign between terms.
(a + b) and (a - b) are conjugates
Examples of conjugates:
(-3 - √2) and (-3 + √2)
(-√3 + 1) and (√3 + 1)
Key fact: (a+b)(a-b) = a² - b² — the middle terms cancel!
✏️ Ex 3 — Simplify (no square root in denominator)
a) -√3 / (√5 - 3)
Multiply by conjugate (√5 + 3)/(√5 + 3)
= (-√3)(√5+3) / (5 - 9) = (-√15 - 3√3) / (-4)
= (√15 + 3√3) / 4
b) 3 / (-1 + 4√5)
Multiply by conjugate (-1 - 4√5)/(-1 - 4√5)
= 3(-1-4√5) / (1 - 80) = (-3 - 12√5) / (-79)
= (3 + 12√5) / 79
c) 2 / (3 + 4√3)
Multiply by (3 - 4√3)/(3 - 4√3)
= 2(3 - 4√3) / (9 - 48) = (6 - 8√3) / (-39)
= (-6 + 8√3) / 39
Cheat Sheet
⚡ Quick Reference
All the Key Formulas
Circle: (x-h)² + (y-k)² = r²
Quadratic Formula: x = (-b ± √(b²-4ac)) / 2a
Vertex Form: y = a(x-h)² + k → vertex = (h, k)
i = √(-1) | i² = -1 | cycle: i, -1, -i, 1 (repeat)
i^n → divide n by 4, use remainder (0→1, 1→i, 2→-1, 3→-i)
Projectile (ft): h = -16t² + vt + s
Projectile (m): h = -4.9t² + vt + s
Discriminant: b²-4ac → (+)=2 real, (0)=1 real, (-)=2 imaginary
Rationalize: multiply top & bottom by the radical in denominator
Trapezoid Area: A = ½(b₁+b₂)h
Perfect Square: a²±2ab+b² = (a±b)²
Difference of Squares: a²-b² = (a+b)(a-b)
Zero Product Property: if AB=0 then A=0 or B=0
Parabola: x=(1/4p)(y-k)²+h or y=(1/4p)(x-h)²+k
Focus (vertical): (h, k+p) | Directrix: y = k-p
Focus (horizontal): (h+p, k) | Directrix: x = h-p