AP PRECALCULUS — UNIT 1B · POLYNOMIAL & RATIONAL FUNCTIONS
1.8 — Rational Functions and Zeros
Notes — Finding x-Intercepts and Solving Rational Inequalities
💡 Learning Objectives
By the end of this lesson you will be able to (AP CED 1.8.A):
- Determine the zeros of a rational function.
- Explain why a zero of the numerator is only a zero of r(x) when the x-value is in the domain.
- Use the zeros and undefined points of r to solve rational inequalities such as r(x) > 0 or r(x) ≤ 0.
1. Where Do Zeros Come From?
A rational function r(x) = N(x) / D(x) equals zero exactly when its numerator equals zero — provided that x is in the domain of r. A fraction equals zero only when its numerator is zero, and its denominator is not zero at that same input.
💡 Definitions
The real zeros of r(x) = N(x)/D(x) are the real values of x for which N(x) = 0 AND D(x) ≠ 0.
- If N(a) = 0 and D(a) ≠ 0, then a is a zero of r.
- If N(a) = 0 and D(a) = 0, then a is NOT automatically a zero of r — we must simplify and check (see Topic 1.10 on holes).
💡 Quick Check
Test your instinct: which x-values are zeros of r(x) = (x − 3)(x + 2) / (x − 5)?
- Candidates from the numerator: x = 3, x = −2.
- Is the denominator nonzero there? Yes — D(3) = −2 ≠ 0, D(−2) = −7 ≠ 0.
So both are zeros. x = 5 is excluded (denominator is zero there).
2. A Worked Example
📘 Example
Example 1 — Finding zeros
Find the real zeros of r(x) = (x² − 4) / (x² − 1).
Numerator factors: x² − 4 = (x − 2)(x + 2), so potential zeros are x = 2 and x = −2.
Denominator: x² − 1 = (x − 1)(x + 1), zero at x = 1 and x = −1.
Neither candidate (2 or −2) makes the denominator zero, so both are zeros of r. x = 1 and x = −1 are excluded from the domain (vertical asymptotes, as we'll see in 1.9).
Zeros of r(x) come from the numerator, provided the x-value is in the domain. The marked points at x = 1 and x = −2 are x-intercepts; the dashed verticals at x = ±3 mark excluded inputs.
⚠️ Common Mistake
Setting the WHOLE rational expression equal to zero and 'cross-multiplying' is an easy way to forget the domain restriction. Always state the excluded values first, then solve N(x) = 0, then check each solution is in the domain.
3. Rational Inequalities
The zeros AND the excluded values of a rational function partition the number line into intervals. On each interval, r(x) has a consistent sign. To solve r(x) > 0 or r(x) < 0, build a sign chart using these boundary points.
📘 Example
Example 2 — Solving a rational inequality with a sign chart
Solve (x − 1)(x + 2) / (x − 3) > 0.
Boundary points: x = 1, x = −2 (zeros of numerator), and x = 3 (excluded). These split the real line into four intervals: (−∞, −2), (−2, 1), (1, 3), (3, ∞).
Test each interval with a sample point: x = −3 gives (−)(−)/(−) = negative; x = 0 gives (−)(+)/(−) = positive; x = 2 gives (+)(+)/(−) = negative; x = 4 gives (+)(+)/(+) = positive.
Solution: (−2, 1) ∪ (3, ∞). Note x = 3 is NOT included (undefined), and x = −2, 1 are NOT included (strict inequality).
🎯 AP Tip
For a non-strict inequality r(x) ≥ 0, INCLUDE the zeros of the numerator (where r equals zero) but NEVER include points where the denominator is zero. The domain always wins.
📘 Try This
Find the zeros of r(x) = (x² − 9) / (x² − 2x − 3).
- Factor numerator: (x − 3)(x + 3).
- Factor denominator: (x − 3)(x + 1).
- Candidates from numerator: x = 3, x = −3.
- Check denominator: at x = 3, denom is also zero — exclude for now. At x = −3, denom = 6 ≠ 0 — keep.
Zero: x = −3 only. (x = 3 is a HOLE, covered in 1.10.)
4. Summary
- Zeros of r(x) = N(x) / D(x) come from N(x) = 0, subject to the domain restriction D(x) ≠ 0.
- Always simplify and check: a common factor between N and D gives a hole, not a zero.
- To solve rational inequalities, combine zeros with excluded inputs to build a sign chart.