AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/1/1-8)
Unit 1 · Topic 1.8
1.8 Rational Functions and Zeros
A rational function equals zero where its numerator is zero, as long as the denominator isn't also zero there. Its zeros, together with the zeros of the denominator, split the number line into intervals where the function keeps one sign, which is how you solve rational inequalities.
Key terms
- zero
- numerator
- domain
- sign chart
- real zero
Finding the zeros
A fraction equals zero only when its top is zero. So the real zeros of r(x) = p(x)/q(x) are the real zeros of p, but only those that are in the domain of r.
The domain excludes every input that makes q(x) = 0. If a number makes both p and q zero, it is not a zero of r; it's a hole or an asymptote (topics 1.9 and 1.10).
Factoring is the cleanest method: factor top and bottom, list the zeros of the top, and cross out any that also zero the bottom.
Where the sign can change
A rational function can switch between positive and negative only at two kinds of inputs: zeros of the numerator and zeros of the denominator. Everywhere else it keeps the same sign.
This is why both sets of zeros are the endpoints, or the asymptote locations, of the intervals that solve r(x) ≥ 0 or r(x) ≤ 0.
Building a sign chart
To solve a rational inequality, make a sign chart:
- Factor the numerator and denominator completely.
- Mark every zero of the numerator and every zero of the denominator on a number line.
- Pick a test input in each interval and find the sign of each factor, then the sign of the whole fraction.
- Choose the intervals with the sign you want.
- Include numerator zeros for ≥ or ≤ (r really equals 0 there). Never include denominator zeros, because r is undefined there.
Connecting to the graph
On the graph, the zeros are the x-intercepts. Intervals where r(x) > 0 are where the graph is above the x-axis, and intervals where r(x) < 0 are where it's below.
A graphing calculator can confirm a sign chart, but on no-calculator questions you need the algebra.
Only real zeros count
Like a polynomial, a numerator can have non-real zeros. Those are not x-intercepts. For example, r(x) = (x² + 1)/(x − 3) has no real zeros, because x² + 1 = 0 only when x = ±i. Its graph never touches the x-axis; it changes sign only at the vertical asymptote x = 3.
Also notice the difference between strict and non-strict inequalities. For r(x) > 0, leave out the zeros of r. For r(x) ≥ 0, put them in. Either way, the inputs that make the denominator zero stay out.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Zeros of a rational function
Find the zeros of r(x) = (x² − x − 6)/(x² − 1).
Show the solutionHide the solution
- Step 1: Factor: r(x) = (x − 3)(x + 2) / ((x − 1)(x + 1)).
- Step 2: Numerator zeros: x = 3 and x = −2.
- Step 3: Neither of these makes the denominator zero (the denominator is zero only at x = 1 and x = −1), so both are in the domain.
Answer: The zeros are x = 3 and x = −2.
- Example 2
Solving a rational inequality
Solve (x² − x − 6)/(x² − 1) ≥ 0.
Show the solutionHide the solution
- Step 1: Boundary values: numerator zeros −2 and 3, denominator zeros −1 and 1. In order: −2, −1, 1, 3.
- Step 2: Test x = −3: (−6)(−1)/((−4)(−2)) = 6/8, positive.
- Step 3: Test x = −1.5: (−4.5)(0.5)/((−2.5)(−0.5)) = −2.25/1.25, negative.
- Step 4: Test x = 0: (−3)(2)/((−1)(1)) = 6, positive.
- Step 5: Test x = 2: (−1)(4)/((1)(3)) = −4/3, negative.
- Step 6: Test x = 4: (1)(6)/((3)(5)) = 6/15, positive.
- Step 7: Keep the positive intervals. Include −2 and 3 (r = 0 there) and exclude −1 and 1 (r is undefined).
Answer: x ≤ −2 or −1 < x < 1 or x ≥ 3, that is, (−∞, −2] ∪ (−1, 1) ∪ [3, ∞).
- Example 3
Trap: a numerator zero that isn't a zero
Find the zeros of f(x) = (x² − 4)/(x² − 2x).
Show the solutionHide the solution
- Step 1: Factor: f(x) = (x − 2)(x + 2) / (x(x − 2)).
- Step 2: The numerator is zero at x = 2 and x = −2.
- Step 3: But x = 2 also makes the denominator zero, so 2 is not in the domain. The function has a hole there, not a zero.
- Step 4: The only zero is x = −2. (Check: x = −2 makes the denominator (−2)(−4) = 8, not zero.)
Answer: The only zero is x = −2.
Common mistakes
- Listing a numerator zero that also makes the denominator zero. Check the domain first.
- Including the denominator's zeros in a ≥ or ≤ solution. The function is undefined there.
- Testing only one interval and assuming the signs alternate. A repeated factor can keep the sign the same across a boundary, so test each interval.
On the exam
- Expect to solve r(x) > 0 or r(x) ≤ 0 by hand, or to identify intervals where a graph is above or below the x-axis.
- Write the solution with clear interval notation and pay attention to brackets versus parentheses.
Connected topics
Videos
Check yourself
4 questions on 1.8 Rational Functions and Zeros. Pick an answer to see if you got it, and why.
What is the solution set of the inequality ((x − 1)(x + 4))/(x − 3) ≥ 0?
Which of the following rational functions has exactly one real zero?
Let r(x) = (2 − x)/((x + 1)(x − 4)). For which values of x is r(x) > 0?
The function r is given by r(x) = (x² − 9)/(x² + x − 6).
What are all the zeros of r?
0 of 4 answered