Topic 1.8 Rational Functions and Zeros: determine the real zeros of a rational function from the zeros of its numerator, accounting for values excluded by the denominator.

What this topic is asking

The College Board (Topic 1.8) wants you to find the real zeros (the $x$-intercepts) of a rational function $f(x) = \frac{p(x)}{q(x)}$. A fraction is zero exactly when its numerator is zero, so the zeros come from $p(x) = 0$, but you must exclude any value that also makes the denominator $q(x)$ zero, because the function is undefined there.

A fraction is zero when its top is zero

This single rule does all the work: solve $p(x) = 0$, then check each solution against $q(x)$.

Excluding the denominator's zeros

A value that makes the denominator zero is never in the domain, so it can never be a zero of the function. If such a value also makes the numerator zero, the shared factor cancels (after which the point becomes a hole, covered in Topic 1.10), but it still is not a zero of $f$. If only the denominator is zero there, the value is a vertical asymptote (Topic 1.9). Either way, you remove it from your zero list.

Multiplicity and crossing behavior

The numerator's factors carry multiplicity just like a polynomial's. After cancelling any factors shared with the denominator, look at the multiplicity of each remaining numerator factor: odd multiplicity makes the graph cross the $x$-axis at that intercept, even multiplicity makes it touch and bounce.

Reading zeros with the rest of the graph

Zeros are one of several features the exam asks you to assemble: zeros (this topic), vertical asymptotes (1.9), holes (1.10) and end behavior (1.7). A clean strategy is to factor numerator and denominator fully first, cancel any shared factors and note the holes they create, then read the remaining numerator factors as zeros and the remaining denominator factors as vertical asymptotes. Doing the factoring once and reading off all the features keeps you from mistaking an excluded value for a zero.

The point that trips students is the difference between "the numerator is zero" and "the function is zero". The numerator being zero is necessary but not sufficient; the denominator must also be nonzero there. Treating the domain restriction as a filter applied after solving the numerator, rather than forgetting it, is the discipline that separates a correct zero list from an incorrect one, and the exam deliberately includes a shared factor to test exactly this.

Try this

Q1. Find the zeros of $f(x) = \frac{x - 7}{x^2 + 4}$. [1 point]

• Cue. Numerator zero at $x = 7$; denominator $x^2 + 4$ is never zero for real $x$, so $x = 7$ is the only zero.

Q2. Does $f(x) = \frac{(x - 1)^2}{x + 2}$ cross or touch the $x$-axis at $x = 1$? [1 point]

• Cue. Multiplicity $2$ (even), so the graph touches and turns back at $x = 1$.