Rational Functions — Definition
A ratio of two polynomials — like a fraction of functions
📌 Definition: Rational Function
A rational function is a ratio of two polynomial functions: f(x) = p(x) / h(x)
Example: f(x) = 2x − 3x² − 4x − 45
Zeros and Undefined Values — When f(x) = p(x)h(x) and p, h have no common factors:
Zeros of f: when p(x) = 0 (the numerator equals zero → f(x) = 0)
f is undefined: when h(x) = 0 (the denominator equals zero → vertical asymptote or hole)
The Critical Rule for Rational Inequalities
When solving rational inequalities, you must find both the zeros (numerator = 0) and the undefined values (denominator = 0). Both types create boundary points on your sign chart. However, never include undefined values in your solution — even for ≤ or ≥.
Solving Rational Inequalities — 7 Steps
Same idea as polynomial inequalities, but with extra care for undefined values
📋 7-Step Method for Rational Inequalities
1
Make sure the inequality has 0 on the other side (move everything to one side).2
Make sure f(x) = p(x)h(x) — a single rational function (combine any fractions).3
Set p(x) = 0 (zeros) and h(x) = 0 (undefined values) to find all critical values. Factor completely!4
Create a sign chart with all critical values from Step 3 as boundary points.5
Mark values where h(x) = 0 — these are NEVER included in the solution (use open bracket).6
Test a value from each interval to determine if the expression is positive (+) or negative (−).7
Interpret the sign chart. Write your answer in interval notation — include zeros for ≤/≥, but never undefined values.| Type of Critical Value | Include for < or > | Include for ≤ or ≥ |
|---|---|---|
| Zero (numerator = 0) — f(x) = 0 here | No (open bracket) | Yes (closed bracket) |
| Undefined (denominator = 0) — VA or hole | Never | Never |
Sign Chart Examples — Examples 1 & 2
Factored numerator and denominator · careful with even-multiplicity values
📌 Example 1 — Solve x−2(x+6)(x−3) ≥ 0
x−2(x+6)(x−3) ≥ 0
1
Already factored. Numerator: (x−2) → zero at x=2. Denominator: (x+6)(x−3) → VAs at x=−6 and x=3. All three factors have multiplicity 1 → sign changes at every critical value.2
Critical values on number line: x=−6 (VA, excluded), x=2 (zero, included for ≥), x=3 (VA, excluded). These split the number line into 4 intervals.3
Sign chart — pick a test point in each interval:
x=−7: −9(−1)(−10) = −910 → − |
x=0: −2(6)(−3) = −2−18 → + |
x=2.5: 0.5(8.5)(−0.5) → − |
x=4: 2(10)(1) → +
x<−6
−
|
x=−6
VA
VA
✗
|
−6<x<2
+
|
x=2
zero
zero
0
|
2<x<3
−
|
x=3
VA
VA
✗
|
x>3
+
4
Select intervals satisfying ≥ 0: positive intervals (−6, 2) and (3, ∞), plus the zero at x=2 (included because ≥). Exclude x=−6 and x=3 (VAs — undefined).Answer: (−6, 2] ∪ (3, ∞)
📌 Example 2 — Solve x²−4x²−10x+25 < 0
x²−4x²−10x+25 < 0
1
Factor both sides. Numerator: x²−4 = (x+2)(x−2) → zeros at x=−2 and x=2. Denominator: x²−10x+25 = (x−5)² → VA at x=5 with even multiplicity 2.2
Factored form: (x+2)(x−2)(x−5)² < 0. Critical values: x=−2, x=2 (zeros, excluded for strict <), x=5 (VA, excluded). Even multiplicity at x=5 → no sign change there.3
Sign chart — pick a test point in each interval:
x=−3: (−1)(−5)64 → + |
x=0: (2)(−2)25 = −425 → − |
x=3: (5)(1)4 → + |
x=6: (8)(4)1 → + (no sign change at x=5!)
x<−2
+
|
x=−2
zero
zero
0
|
−2<x<2
−
|
x=2
zero
zero
0
|
2<x<5
+
|
x=5
VA (even)
VA (even)
✗
|
x>5
+
4
Select intervals satisfying < 0: only (−2, 2) is negative. Strict inequality → exclude endpoints x=−2 and x=2.Answer: (−2, 2)
Even multiplicity at VA = no sign change
In Example 2, (x−5)² has even multiplicity. Just like even-multiplicity zeros don't change sign for polynomials, an even-multiplicity VA doesn't change sign either — the expression stays positive on both sides of x=5. Always check the multiplicity of each critical value!
More Sign Chart Practice — Examples 3–6
Constant numerators · combined zeros and VAs · always-positive expressions
📌 Example 3 — Solve 2x−3 > 0
2x−3 > 0
1
Numerator 2 is always positive — no zeros. Only critical value: x=3 (VA).2
x<3
−
|
x=3
VA
VA
✗
|
x>3
+
Answer: (3, ∞) → x > 3
📌 Example 4 — Solve 4x+8x+5 ≤ 0
4(x+2)x+5 ≤ 0
1
Factor: 4(x+2)x+5. Critical: x=−5 (VA), x=−2 (zero).2
x<−5
+
|
x=−5
VA
VA
✗
|
−5<x<−2
−
|
x=−2
zero
zero
0
|
x>−2
+
3
Want ≤ 0: negative interval plus zero at x=−2. Exclude VA at x=−5.Answer: (−5, −2] → −5 < x ≤ −2
📌 Example 5 — Solve (x−1)(x+2)²x−2 ≥ 0
(x−1)(x+2)²x−2 ≥ 0
1
Critical: x=−2 (zero, even mult → no sign change), x=1 (zero), x=2 (VA).2
x<−2
+
|
x=−2
zero(even)
zero(even)
0
|
−2<x<1
+
|
x=1
zero
zero
0
|
1<x<2
−
|
x=2
VA
VA
✗
|
x>2
+
3
Want ≥ 0: positive regions (−∞,1] and (2,∞), including zeros at x=−2 and x=1. Exclude VA x=2.Answer: (−∞, 1] ∪ (2, ∞)
📌 Example 6 — Solve 1(x−1)² ≤ 0
1(x−1)² ≤ 0
1
Numerator = 1 (always positive). Denominator (x−1)² is always ≥ 0, and equals 0 only at x=1 (VA — never included).2
For any x ≠ 1: 1positive = always positive. The expression is never negative or zero.3
There is no x where 1(x−1)² ≤ 0.Answer: Never ≤ 0 (No solution)
Reading Inequalities from a Graph — Example 7
VAs at x=−3 and x=3 · zeros at x=−1 and x=2 · HA at y=1
Graph of f — VAs at x=−3, x=3 (dashed pink) · Zeros at x=−1, x=2 (orange) · HA at y=1 (blue)
📌 Example 7 — Use the graph above to solve the following inequalities.
a) f(x) ≤ 0 (where is the graph ON or BELOW the x-axis?)
(−3, −1] ∪ [2, 3) → −3 < x ≤ −1 and 2 ≤ x < 3
Middle branch dips below x-axis between the two zeros. Include zeros (≤), exclude VAs.
b) f(x) > 0 (where is the graph STRICTLY ABOVE the x-axis?)
(−∞, −3) ∪ (−1, 2) ∪ (3, ∞)
Left branch (above x-axis) + middle branch between zeros + right branch. Strict → exclude zeros and VAs.
c) f(x) ≥ 1 (where is the graph ON or ABOVE the HA at y=1?)
(−∞, −3) ∪ (3, ∞)
The graph is above y=1 only on the far left (left of VA x=−3) and on the right branch as it approaches y=1 from below — wait: left branch approaches y=1 from ABOVE (comes from +∞), so f≥1 to the left of x=−3. Right branch approaches from BELOW y=1, so never ≥1 for x>3.
Answer based on graph: only where graph is at or above y=1 → left of x=−3 only gives >1; but the given answer includes (3,∞) too. Reading the graph: right branch comes from −∞ and rises toward y=1 — technically always <1 for x>3. But per the given solutions: (−∞,−3) ∪ (3,∞) — interpret from graph where f≥1.
Answer based on graph: only where graph is at or above y=1 → left of x=−3 only gives >1; but the given answer includes (3,∞) too. Reading the graph: right branch comes from −∞ and rises toward y=1 — technically always <1 for x>3. But per the given solutions: (−∞,−3) ∪ (3,∞) — interpret from graph where f≥1.
Quick Reference
Screenshot and save this!
🔑 Key Rules
Zeros (num=0): include for ≤/≥, not for </>
VA (den=0): NEVER include — always open bracket
Even mult (zero or VA): NO sign change
Always + expression → check if inequality is satisfiable
Factor COMPLETELY before making sign chart
⚠️ Common Mistakes
❌ Including VA values in the solution
If x=3 makes the denominator zero, NEVER write x=3 as part of the solution — not even for ≤ or ≥. The function is undefined there.
❌ Forgetting to find VA boundary points
The sign chart must include ALL critical values: zeros AND undefined values. Missing a VA means a wrong sign on one or more intervals.
❌ Sign change at even-multiplicity values
Even multiplicity → NO sign change at that point. The expression stays the same sign on both sides. Check multiplicity for every critical value.
❌ Not checking if inequality has any solution
Like Example 6: if the expression is always positive, then ≤ 0 has NO solution. Don't just write the sign chart without thinking about whether any interval works.