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📚 Topic 1.7 · Unit 1

Rational Functions &
End Behavior

Horizontal asymptotes, slant asymptotes, and limit statements — determined by comparing degrees.

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📖
What is a Rational Function?
The definition and basic form
f(x) = p(x) / q(x)
A rational function is the quotient of two polynomials, where q(x) ≠ 0.
n = degree of p(x)
n is the degree of the numerator. The leading term is axⁿ.
d = degree of q(x)
d is the degree of the denominator. The leading term is bx^d.
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End behavior is determined by leading terms only

As x → ±∞, lower-degree terms become negligible. Only the leading terms of the numerator and denominator matter:

f(x) ≈ (axⁿ) / (bx^d) as x → ±∞

The relationship between n and d tells you everything about end behavior.

⚖️
The Three Cases
Compare n (numerator degree) vs d (denominator degree)
Case 1
n = d
Same degree → Horizontal Asymptote at y = a/b (ratio of leading coefficients)
3x²/5x² → y = 3/5
Case 2
n < d
Denominator dominates → Horizontal Asymptote at y = 0
2x/x² → y = 0
Case 3
n > d
Numerator dominates → No horizontal asymptote. End behavior like y = (a/b)x^(n−d)
x³/x² → behaves like y = x
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Special Case: Slant (Oblique) Asymptote

When n = d + 1 (numerator degree is exactly 1 more than denominator), the function has a slant asymptote.

The slant asymptote is parallel to the line y = (a/b)x where a and b are the leading coefficients.

⚠️ If n = d + 2 or more, there is no slant asymptote — the numerator dominates too much.

✏️
Example 1 — Identify Asymptote Type
Horizontal, slant, or neither?
f(x) = (3x²+4x−7) / (5x²−3)
n=2, d=2 → n=d → Case 1
Leading coefficients: 3 and 5
HA: y = 3/5
y = (2x−5) / (x²+3x+2)
n=1, d=2 → n<d → Case 2
Denominator dominates
HA: y = 0
f(x) = (2x²−4) / (5x+9)
n=2, d=1 → n=d+1
Slant! Leading ratio: 2/5
Slant asymptote (parallel to y = 2/5 x)
y = (4x+5) / (8x−1)
n=1, d=1 → n=d → Case 1
Leading coefficients: 4 and 8
HA: y = 4/8 = 1/2
f(x) = 3 / (x²+3x−7)
n=0, d=2 → n<d → Case 2
Denominator dominates
HA: y = 0
f(x) = −4 / (2x+1)
n=0, d=1 → n<d → Case 2
Denominator dominates
HA: y = 0
↔️
Writing Limit Statements
Example 2 — describing end behavior formally
🎯

How to write limit statements

Identify the case, then determine what value (or ±∞) the function approaches:

• Case 1 (n=d): both limits = a/b
• Case 2 (n<d): both limits = 0
• Case 3 (n>d): use end behavior of (a/b)x^(n−d) — check sign carefully for x→−∞

📌 Example 2a — f(x) = (2x³+4x−1) / (6x³−x²+4)
1
n=3, d=3 → n=d → Case 1. Leading coefficients: 2 and 6.
2
Ratio = 2/6 = 1/3. Horizontal asymptote y = 1/3.
3
Both limits equal the HA value.
lim x→−∞ f(x) = 1/3    lim x→+∞ f(x) = 1/3
📌 Example 2b — g(x) = (5x²−8x+9) / (2x³+x−1)
1
n=2, d=3 → n<d → Case 2. Denominator dominates.
lim x→−∞ g(x) = 0    lim x→+∞ g(x) = 0
📌 Example 2c — h(x) = (−3x⁴−x²+x) / (x³+4x+4)
1
n=4, d=3 → n>d → Case 3. End behavior like (−3/1)x^(4−3) = −3x.
2
As x→−∞: −3x → −3(−∞) = +∞
3
As x→+∞: −3x → −3(+∞) = −∞
lim x→−∞ h(x) = +∞    lim x→+∞ h(x) = −∞
CaseConditionlim x→−∞lim x→+∞
1n = da/ba/b
2n < d00
3 (odd n−d, a/b > 0)n > d−∞+∞
3 (odd n−d, a/b < 0)n > d+∞−∞
3 (even n−d, a/b > 0)n > d+∞+∞
3 (even n−d, a/b < 0)n > d−∞−∞
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Slant Asymptotes — Example 3
n = d + 1 exactly
⚠️

The Exact Requirement

A slant asymptote exists only when n = d + 1 — the numerator degree is exactly 1 more.

The slant asymptote is parallel to the line y = (a/b)x, where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator.

If n = d + 2 or more → no slant, no horizontal asymptote (numerator dominates too much).

📌 Example 3 — Which function has a slant asymptote parallel to y = (1/2)x?
I
f(x) = (x²+3)/(2x²+x+6): n=2, d=2 → n=d → horizontal asymptote, not slant. ✗
II
g(x) = (x²+4x+1)/(2x³+x²+2): n=2, d=3 → n<d → HA y=0, not slant. ✗
III
h(x) = (x²+3x+2)/(2x+4): n=2, d=1 → n=d+1 ✓ → slant. Leading ratio = 1/2. Slant parallel to y=(1/2)x ✓
IV
k(x) = (x⁴+x³+5)/(2x²+x−1): n=4, d=2 → n=d+2 → NOT slant (need exactly n=d+1). ✗
Answer: C) III only — h(x) has slant asymptote parallel to y = (1/2)x ✓
⚡
Quick Reference
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🔑 Decision Tree
n = d → HA: y = a/b
Same degree → ratio of leading coefficients
n < d → HA: y = 0
Denominator dominates → approaches zero
n = d+1 → Slant: y = (a/b)x
Exactly 1 more → slant asymptote
n > d+1 → No asymptote
Numerator dominates → end behavior like polynomial
⚠️ Common Mistakes
❌ HA = ratio of ALL coefficients
Only the leading coefficients matter for end behavior. Ignore all other terms.
❌ n>d always means slant asymptote
Slant requires exactly n=d+1. If n=d+2 or more, there is no horizontal or slant asymptote.
❌ Sign doesn't matter for limits
In Case 3, always check the sign of a/b when substituting x→−∞ — a negative leading coefficient flips the sign.
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