🏠 Homeβ€Ί AP Precalculusβ€Ί Unit 1β€Ί 1.11 Notes
πŸ“ Topic 1.11 Β· Unit 1

Equivalent Representations of
Polynomial & Rational Functions

Holes Β· Vertical Asymptotes Β· Slant Asymptotes Β· Long Division Β· Pascal's Triangle Β· Binomial Theorem

πŸ“‹ Notes πŸƒ Flashcards 🧠 Quiz
πŸ“Œ Recall β€” Holes vs Vertical Asymptotes
πŸ”
The Key Rule
Both come from zeros of the denominator β€” the difference is whether the numerator shares that zero
πŸ“‹

The Recall

Let f(x)g(x) = h(x). If g(c) = 0, then h(x) has either a vertical asymptote or a hole at x = c.

πŸ•³οΈ Hole vs Vertical Asymptote β€” What's the Difference?
πŸ•³οΈ Hole at x = c
When: (x βˆ’ c) is a factor of BOTH numerator AND denominator (common zero β€” it cancels)
Result: The factor cancels out. The function is defined everywhere except x = c β€” it has a removable discontinuity (open circle on graph)
⚑ Vertical Asymptote at x = c
When: (x βˆ’ c) is a factor of the denominator ONLY β€” not in the numerator
Result: The function blows up to ±∞ as x β†’ c. The graph has a vertical line it never crosses.
πŸ“Œ Example 1 β€” Let h(x) = xΒ² βˆ’ 4xΒ² + 7x + 10. Write in factored form. Find any holes or vertical asymptotes.
1
Factor numerator: xΒ² βˆ’ 4 = (x βˆ’ 2)(x + 2)
2
Factor denominator: xΒ² + 7x + 10 = (x + 2)(x + 5)
3
Simplify: h(x) = (x βˆ’ 2)(x + 2)(x + 2)(x + 5) = x βˆ’ 2x + 5
πŸ•³οΈ
Hole at x = βˆ’2: (x + 2) is a common factor β€” it cancels, creating a hole.
⚑
Vertical asymptote at x = βˆ’5: (x + 5) remains in denominator only β€” it does NOT cancel.
Factored: h(x) = (x βˆ’ 2)/(x + 5)  Β·  Hole: x = βˆ’2  Β·  VA: x = βˆ’5
πŸ“Œ Example 2 β€” Find the domain of h(x) from Example 1.
1
Exclude ALL values where the original denominator = 0: x = βˆ’2 (hole) and x = βˆ’5 (VA)
2
Write in interval notation, removing both excluded values from all real numbers
Domain: (βˆ’βˆž, βˆ’5) βˆͺ (βˆ’5, βˆ’2) βˆͺ (βˆ’2, ∞)
πŸ“Œ Example 3 β€” Let k(x) = xΒ² βˆ’ x βˆ’ 12xΒ³ + xΒ² βˆ’ 20x. Full analysis.
a
Factor: k(x) = (x βˆ’ 4)(x + 3)x(x βˆ’ 4)(x + 5) = x + 3x(x + 5)
b
Zeros: x = βˆ’3, because (x + 3) is in the numerator only (not denominator)
c
Hole: x = 4, because (x βˆ’ 4) is a common factor β€” cancels
d
Vertical asymptotes: x = βˆ’5 and x = 0 β€” denominator factors that don't cancel
e
Horizontal asymptote: y = 0 β€” degree of denominator (3) > degree of numerator (2)
f
Domain: (βˆ’βˆž, βˆ’5) βˆͺ (βˆ’5, 0) βˆͺ (0, 4) βˆͺ (4, ∞)
πŸ“ˆ Reading Rational Graphs β€” Examples 4 & 5
πŸ“ˆ
Building the Equation from a Graph
Read holes, zeros, VAs, and HA from the graph β†’ write the factored equation
πŸ—ΊοΈ

Strategy β€” what each graph feature tells you

Open circle (hole) at x = a β†’ factor (x βˆ’ a) in both numerator AND denominator
Zero / x-intercept at x = a β†’ factor (x βˆ’ a) in numerator only
Vertical asymptote at x = a β†’ factor (x βˆ’ a) in denominator only
Horizontal asymptote y = k β†’ if same degree: ratio of leading coefficients = k. If denom higher degree: y = 0

πŸ“Œ Example 4 β€” Graph shows: Hole at x = 4, Zero at x = 3, VA at x = 2, HA at y = βˆ’1
1
Hole x = 4 β†’ (x βˆ’ 4) in both top and bottom
2
Zero x = 3 β†’ (x βˆ’ 3) in numerator only
3
VA x = 2 β†’ (x βˆ’ 2) in denominator only
4
HA y = βˆ’1 β†’ same degree, leading ratio = βˆ’1
f(x) = βˆ’(x βˆ’ 4)(x βˆ’ 3)(x βˆ’ 4)(x βˆ’ 2)
πŸ“Œ Example 5 β€” Graph shows: Hole at x = 2, Zero at x = 1, VAs at x = 0 and x = 3, HA at y = 0, approaches 0⁺ from left
1
Hole x = 2 β†’ (x βˆ’ 2) in both
2
Zero x = 1 β†’ (x βˆ’ 1) in numerator
3
VAs x = 0, 3 β†’ x(x βˆ’ 3) in denominator
4
HA y = 0, approaches 0⁺ β†’ leading ratio = βˆ’1
g(x) = βˆ’(x βˆ’ 2)(x βˆ’ 1)x(x βˆ’ 2)(x βˆ’ 3)
πŸ“ Slant Asymptotes & Long Division
πŸ“
When Degree of Numerator = Denominator + 1
No horizontal asymptote β€” instead a slant (oblique) asymptote found by long division
πŸ’‘

Asymptote Type β€” Quick Reference

deg(num) < deg(denom): Horizontal asymptote y = 0
deg(num) = deg(denom): Horizontal asymptote y = (ratio of leading coefficients)
deg(num) = deg(denom) + 1: Slant asymptote β€” use long division to find it
deg(num) > deg(denom) + 1: No horizontal or slant asymptote

πŸ”’ Numerical Long Division
32 R4 21 ) 676 βˆ’63 β€”β€”β€” 46 βˆ’42 β€”β€”β€” 4
Quotient: 32   Remainder: 4
Answer: 32 + 421 = 32421
πŸ“ Polynomial Long Division
3x + 2 2x+1 ) 6xΒ² + 7x + 6 βˆ’(6xΒ² + 3x) β€”β€”β€”β€”β€”β€”β€”β€” 4x + 6 βˆ’(4x + 2) β€”β€”β€”β€”β€”β€”β€”β€” 4
Quotient: 3x + 2   Remainder: 4
Answer: (3x + 2) + 42x + 1
πŸ“Œ Example 6 β€” Let h(x) = 6xΒ² + x + 52x + 1. Use long division to find the slant asymptote.
1
Degree of numerator (2) = degree of denominator (1) + 1 β†’ slant asymptote exists
2
Perform long division:
3x βˆ’ 1 2x+1 ) 6xΒ² + x + 5 βˆ’(6xΒ² + 3x) β€”β€”β€”β€”β€”β€”β€”β€” βˆ’2x + 5 βˆ’(βˆ’2x βˆ’ 1) β€”β€”β€”β€”β€”β€”β€”β€” 6
3
h(x) = (3x βˆ’ 1) + 62x + 1 β†’ as x β†’ ±∞, the fraction β†’ 0, so the slant asymptote is the quotient only
Slant asymptote: y = 3x βˆ’ 1
πŸ“Œ Example 7 β€” Let g(x) = xΒ³ + 4xΒ² βˆ’ 12xxΒ² + 7x + 6. Find ALL asymptotes.
1
Factor: g(x) = x(x βˆ’ 2)(x + 6)(x + 1)(x + 6) = x(x βˆ’ 2)x + 1
2
Vertical asymptote: x = βˆ’1 (denominator zero not in numerator). Hole: x = βˆ’6 (common factor)
3
Horizontal asymptote: None β€” degree of numerator (3) > degree of denominator (2)
4
Slant asymptote (deg(num) = deg(denom) + 1): divide xΒ² βˆ’ 2x by x + 1
x βˆ’ 3 x+1 ) xΒ² βˆ’ 2x βˆ’(xΒ² + x) β€”β€”β€”β€”β€”β€”β€”β€” βˆ’3x βˆ’(βˆ’3x βˆ’ 3) β€”β€”β€”β€”β€”β€”β€”β€” 3
Vertical asymptote: x = βˆ’1  Β·  Hole: x = βˆ’6  Β·  Slant asymptote: y = x βˆ’ 3
πŸ“Œ Example 8 β€” Let d(x) = 2x(x + 4)(x βˆ’ 1)xΒ² βˆ’ 2x + 1. Which statement is correct?
1
Factor denominator: xΒ² βˆ’ 2x + 1 = (x βˆ’ 1)Β²   β†’   d(x) = 2x(x + 4)(x βˆ’ 1)(x βˆ’ 1)Β² = 2x(x + 4)x βˆ’ 1
2
Hole? (x βˆ’ 1) cancels once β†’ hole at x = 1 (not VA). One remaining (x βˆ’ 1) in denom β†’ VA at x = 1. Wait β€” (x βˆ’ 1) appears once in numerator, twice in denominator β†’ cancels ONCE, leaving one in denom β†’ VA at x = 1, hole = none
3
Asymptote type: deg(num)=3, deg(denom)=2, difference=1 β†’ slant asymptote. Leading coefficient ratio β†’ slant y = 2x
(A)
1 HA, 1 hole, 1 VA
No HA (slant asymptote instead) βœ—
(B)
1 HA, 2 VAs
No HA, only 1 VA βœ—
(C)
1 slant, 1 hole, 1 VA
No hole β€” (xβˆ’1) multiplicity: denom has 2, numerator has 1, net = 1 still in denom βœ—
βœ“ (D)
1 slant, no holes, 1 VA
Slant asymptote y=2x, VA at x=1, no holes βœ“
πŸ”Ί Pascal's Triangle & The Binomial Theorem
πŸ”Ί
Expanding (a + b)ⁿ Without Multiplying Everything Out
Pascal's Triangle gives the coefficients Β· powers follow a strict pattern
πŸ“‹

The Binomial Theorem

(a+b)ⁿ = C(n,0)aⁿ + C(n,1)aⁿ⁻¹b + C(n,2)aⁿ⁻²bΒ² + … + C(n,n)bⁿ

Three key patterns to remember:
1. "a" starts at degree n, decreases by 1 each term
2. "b" starts at degree 0, increases by 1 each term
3. Coefficients come from Pascal's Triangle (row n)

Pascal's Triangle
Row 0 β†’
1
Row 1 β†’
11
Row 2 β†’
121
Row 3 β†’
1331
Row 4 β†’
14641
Row 5 β†’
15101051
Row 6 β†’
1615201561
Row 7 β†’
172135352171
The circled 10 above is the 2nd element of Row 5 β†’ C(5,2) = 10. First number in each row is the 0th element!
πŸ”‘

C(n, r) β€” "n choose r"

The coefficient C(n,r) = n!r!(nβˆ’r)! gives the rth element of row n in Pascal's Triangle. You don't need the formula for AP Precalc β€” just read from the triangle. Also written as ⁿCα΅£ or C(n,r).

πŸ“Œ Example 10 β€” Use Pascal's Triangle to expand (x + 2)⁡.
1
Use Row 5: coefficients are 1, 5, 10, 10, 5, 1
2
Apply: a = x (powers 5β†’0), b = 2 (powers 0β†’5)
1·x⁡ + 5·x⁴·2¹ + 10·x³·2² + 10·x²·2³ + 5·x·2⁴ + 1·2⁡
3
Simplify each term:
x⁡ + 10x⁴ + 40x³ + 80x² + 80x + 32
(x + 2)⁡ = x⁡ + 10x⁴ + 40x³ + 80x² + 80x + 32
πŸ“Œ Example 11 β€” Use Pascal's Triangle to expand (2x βˆ’ 1)⁴.
1
Use Row 4: coefficients are 1, 4, 6, 4, 1
2
a = 2x, b = βˆ’1 (keep the negative with b)
(2x)⁴ + 4(2x)Β³(βˆ’1) + 6(2x)Β²(βˆ’1)Β² + 4(2x)(βˆ’1)Β³ + (βˆ’1)⁴
3
Compute:
16x⁴ + 4(8xΒ³)(βˆ’1) + 6(4xΒ²)(1) + 4(2x)(βˆ’1) + 1
= 16x⁴ βˆ’ 32xΒ³ + 24xΒ² βˆ’ 8x + 1
(2x βˆ’ 1)⁴ = 16x⁴ βˆ’ 32xΒ³ + 24xΒ² βˆ’ 8x + 1
πŸ“Œ Example 12 β€” What is the coefficient of x⁴ when (x + 5)⁢ is expanded?
1
The x⁴ term uses Row 6, element 2 (since b appears twice when a=x⁴): C(6,2) = 15
2
Term: C(6,2) · x⁴ · 5² = 15 · x⁴ · 25 = 375x⁴
Coefficient of x⁴ = 375
πŸ“Œ Example 13 β€” What is the coefficient of xΒ³ when (x βˆ’ 3)⁸ is expanded?
1
xΒ³ term: a = x appears 3 times β†’ b = (βˆ’3) appears 5 times. Coefficient = C(8,5)
2
C(8,5) = 8!5! Β· 3! = 8 Β· 7 Β· 66 = 56
3
Term: 56 Β· xΒ³ Β· (βˆ’3)⁡ = 56 Β· xΒ³ Β· (βˆ’243) = βˆ’13,608xΒ³
Coefficient of xΒ³ = βˆ’13,608
⚑ Quick Reference β€” Asymptote Rules
SituationTypeHow to FindExample
deg(num) < deg(denom) Horizontal y = 0 Automatically y = 0 x + 1xΒ² + 3 β†’ y = 0
deg(num) = deg(denom) Horizontal y = k Ratio of leading coefficients 3xΒ²2xΒ² β†’ y = 32
deg(num) = deg(denom) + 1 Slant y = mx + b Long division β€” quotient is the asymptote 6xΒ² + x + 52x + 1 β†’ y = 3x βˆ’ 1
Denom factor cancels with numer Hole (open circle) Set cancelled factor = 0 (x+2) cancels β†’ hole at x = βˆ’2
Denom factor doesn't cancel Vertical asymptote Set remaining denom factor = 0 (x+5) stays β†’ VA at x = βˆ’5