Unit 1 · Pre-Calculus
Polynomial &
Rational Functions
From domain and range to asymptotes and inequalities — master every core concept with clear explanations and worked examples.
① Characteristics of Functions
② Polynomial Functions
③ Remainder & Factor Theorems
④ Rational Functions
⑤ Polynomial & Rational Inequalities
Domain
All valid inputs
The set of all x-values you can plug into f(x). Always watch for: denominator ≠ 0 and radicand ≥ 0.
Range
All possible outputs
The set of all y-values the function actually produces. It's a subset of the codomain — read it off the graph vertically.
Intervals of Increase / Decrease / Constant
How the output moves as x grows
As x increases left → right:
• f(x) rises → increasing
• f(x) falls → decreasing
• f(x) stays flat → constant
Split intervals at turning points and always write them in x-values only (not y).
📝 Example 1
Find the domain and range of f(x) = √(x − 3) + 1.
✦ Solution
1
The expression under the radical must be ≥ 0: x − 3 ≥ 0 → x ≥ 3
3
When x ≥ 3, we get √(x−3) ≥ 0, so f(x) = √(x−3) + 1 ≥ 1
AnswerDomain: [3, ∞) | Range: [1, ∞)
💡
Quick checklist for domain restrictions: ① denominator = 0, ② radicand < 0, ③ log argument ≤ 0. Eliminate those x-values and you're done.
End Behavior
Where does the graph go at the far left and right?
Determined entirely by the leading coefficient and the degree. Everything else in the middle doesn't matter for the ends.
| Degree | Leading coeff. | x → −∞ | x → +∞ | Shape |
|---|
| Even | Positive (+) | +∞ | +∞ | U-shape ↗↗ |
| Even | Negative (−) | −∞ | −∞ | ∩-shape ↘↘ |
| Odd | Positive (+) | −∞ | +∞ | Rising left to right / |
| Odd | Negative (−) | +∞ | −∞ | Falling left to right \ |
Zeros & Multiplicity
How the graph behaves at each x-intercept
A factor (x − r)^k gives zero x = r with multiplicity k.
• k is odd → graph crosses the x-axis (sign changes)
• k is even → graph touches the x-axis and bounces back (sign stays)
📝 Example 2
For f(x) = −2(x+1)²(x−3), state the end behavior, zeros, and each multiplicity.
✦ Solution
1
Degree 3, leading coefficient −2 (odd degree, negative) → as x→−∞, f→+∞; as x→+∞, f→−∞
2
Zero at x = −1 with multiplicity 2 → graph touches and bounces
3
Zero at x = 3 with multiplicity 1 → graph crosses the axis
4
y-intercept: f(0) = −2(1)²(−3) = 6
AnswerEnd: ↗ then ↘ | x=−1 (touch), x=3 (cross) | y-int = 6
🎯
Degree check: The sum of all multiplicities always equals the degree of the polynomial.
Remainder Theorem
Find remainders instantly — no long division
When polynomial P(x) is divided by (x − a), the remainder equals P(a).
Just plug in the value and evaluate!
Factor Theorem
Test whether (x − a) is a factor in one step
P(a) = 0 ⟺ (x − a) is a factor of P(x).
It's the Remainder Theorem with remainder = 0. Zero remainder means exact factor.
📝 Example 3
Fully factor P(x) = x³ − 4x² + x + 6.
✦ Solution
1
Test factors of the constant term 6: try ±1, ±2, ±3, ±6
2
P(2) = 8 − 16 + 2 + 6 = 0 ✓ → (x − 2) is a factor
3
Use synthetic division to divide out (x − 2): quotient is x² − 2x − 3
4
Factor the quotient: x² − 2x − 3 = (x − 3)(x + 1)
AnswerP(x) = (x − 2)(x − 3)(x + 1)
⚡
Strategy: Test candidates → confirm zero → synthetic division → factor quotient → repeat until fully factored.
Vertical Asymptote (VA)
Where the denominator = 0
After canceling common factors, the zeros of the remaining denominator are vertical asymptotes. The graph approaches but never crosses.
Horizontal Asymptote (HA)
Compare degrees of numerator vs. denominator
deg(num) < deg(den) → y = 0
deg(num) = deg(den) → y = ratio of leading coefficients
deg(num) > deg(den) → no horizontal asymptote
Oblique (Slant) Asymptote
When deg(num) = deg(den) + 1
Perform polynomial long division. The quotient (ignoring the remainder) is the oblique asymptote y = ax + b.
Example: (x² + 1) ÷ (x − 1) → quotient x + 1, remainder 2 → asymptote is y = x + 1
📝 Example 4
Find all asymptotes of f(x) = (2x² − x − 3) / (x² − 4).
✦ Solution
1
Factor numerator: 2x² − x − 3 = (2x − 3)(x + 1)
2
Factor denominator: x² − 4 = (x − 2)(x + 2). No common factors → no holes.
3
Vertical asymptotes: x = 2 and x = −2
4
Degrees equal (both 2nd) → horizontal asymptote y = 2/1 = 2
AnswerVA: x = 2, x = −2 | HA: y = 2 | No oblique
| Condition | Type | How to find it |
|---|
| Denom = 0 (can't cancel) | Vertical asymptote | Zeros of reduced denominator |
| deg(num) < deg(den) | Horizontal y = 0 | Limit as x → ∞ |
| deg(num) = deg(den) | Horizontal y = a/b | Ratio of leading coefficients |
| deg(num) = deg(den) + 1 | Oblique y = ax + b | Quotient of long division |
| deg(num) ≥ deg(den) + 2 | No asymptote | — |
Core Strategy
The Sign Chart Method
① Move everything to one side (right side = 0)
② Fully factor the expression (for rationals: combine into one fraction, then factor)
③ Find critical values — zeros and excluded points (where denom = 0)
④ Plot critical values on a number line; test the sign in each interval
⑤ Select intervals matching your inequality direction
⚠️
Rational inequality warning: Never multiply both sides by the denominator if you don't know its sign. Always use the sign chart instead.
📝 Example 5
Solve (x − 1) / (x + 2) ≥ 0.
✦ Solution
1
Critical values: x = 1 (numerator = 0), x = −2 (denominator = 0, excluded)
Interval | x < −2 | −2 < x < 1 | x > 1
──────────────────────────────────────────────
(x − 1) | (−) | (−) | (+)
(x + 2) | (−) | (+) | (+)
Overall | (+) | (−) | (+)
3
We need ≥ 0 → take positive intervals plus where numerator = 0. Exclude x = −2.
Answerx < −2 or x ≥ 1
📝 Example 6
Solve the polynomial inequality x³ − x² − 6x > 0.
✦ Solution
1
Factor: x(x² − x − 6) = x(x − 3)(x + 2)
2
Critical values (zeros): x = −2, 0, 3
3
Each zero has multiplicity 1 (odd) → sign flips at every critical value:
x < −2: (−) | −2 < x < 0: (+) | 0 < x < 3: (−) | x > 3: (+)
4
We need > 0 → take the positive intervals (strict, so endpoints excluded)
Answer−2 < x < 0 or x > 3
🏆
Inequality checklist: Move to one side → factor → critical values on number line → sign chart → select correct intervals → check whether endpoints are included or excluded.