A zero of a function is an input that makes the output 0. For a rational function r(x) = N(x) ÷ D(x), a fraction is 0 only when its top is 0. So the real zeros of r correspond to the real zeros of the numerator, for the values that are in the domain of r.
To find them, set the numerator equal to 0 and solve. Factoring the numerator makes this quick. For r(x) = (x − 2)(x + 3)/(x − 4), the numerator is 0 at x = 2 and x = −3. Those are the x-intercepts.
A fraction equals zero only when its top is zero, so the real zeros of r are the real zeros of the numerator. Green dots are the x-intercepts. A dashed orange line marks a zero of the denominator, where r is not defined.
The phrase "for such values in its domain" matters. An input is a zero of r only if two things are true: the numerator is 0 and the denominator is not 0. If an input makes the denominator 0, it is not in the domain, so it cannot be a zero, even if the numerator is also 0 there.
Try r(x) = (x² − 4)/(x + 2). The numerator is 0 at x = 2 and x = −2. But at x = −2 the denominator is also 0, so −2 is not in the domain. The only real zero is x = 2. (What the graph does near −2 is a topic for a later lesson.)
Two checks for any candidate input a: is the denominator nonzero (so a is in the domain), and is the numerator zero? Pick a function and a candidate value.
| Numerator N(2) | Denominator D(2) | In domain? | Zero of r? |
|---|---|---|---|
| 0 | −2 | yes | yes |
To solve an inequality like r(x) ≥ 0 or r(x) ≤ 0 you need to know where r is positive and where it is negative. The real zeros of both polynomials of the rational function r are the places where that can change. They are endpoints or asymptotes for the intervals that satisfy the inequality: a zero of the numerator is an endpoint (the graph crosses or touches the x-axis), and a zero of the denominator is an asymptote (the graph is not defined there).
Call all of these critical values. Mark them on a number line. k critical values make k + 1 intervals. On each interval r(x) never changes sign, so one test point tells you the sign of the whole interval.
The real zeros of the numerator and of the denominator are the critical values. They cut the number line into intervals, and r(x) keeps one sign on each interval. Pick a test point in each interval to find that sign.
| interval | test point | value there | sign |
|---|---|---|---|
| (−∞, −2) | −3 | −0.667 | − |
| (−2, 1) | 0 | 0.667 | + |
| (1, 3) | 2 | −4 | − |
| (3, ∞) | 4 | 18 | + |
Once you have the signs, read off the intervals. For r(x) > 0 take the intervals marked +; for r(x) < 0 take the intervals marked −. For ≥ or ≤ you may also include the endpoints that are zeros of the numerator, because there r(x) = 0. You may never include a zero of the denominator.
| Critical value | With > or < | With ≥ or ≤ |
|---|---|---|
| zero of the numerator | excluded | included |
| zero of the denominator | excluded | excluded |
Green marks where the inequality is true. Switch between strict and non-strict inequalities and watch the endpoints: a zero of the numerator joins the solution with ≥ or ≤, but a zero of the denominator never does.
Multiplicity from Lesson 1.5 controls whether the sign flips. At a critical value of odd multiplicity the sign changes. At a critical value of even multiplicity it does not. This is true for zeros of the numerator and for zeros of the denominator.
So r(x) = (x + 1)/(x − 2)² is positive on both sides of x = 2, and r(x) = (x − 3)²/(x + 1) is positive on both sides of x = 3. When you test points, a double zero tells you to expect the same sign on both sides. It is also a good way to check your work.
r(x) = (x − 1)a / (x + 2)b. An odd multiplicity makes the sign change at a critical value. An even multiplicity does not. Choose a and b.
| interval | sign of r |
|---|---|
| (−∞, −2) | + |
| (−2, 1) | − |
| (1, ∞) | + |