Rational Functions and Zeros
Introduced| English |
|---|
| rational function/ˈræʃənl ˈfʌŋkʃn/ |
| zeros/ˈzɪərəʊz/ |
| numerator/ˈnjuːməreɪtə/ |
| domain/dəˈmeɪn/ |
| denominator/dɪˈnɒmɪneɪtə/ |
When is a fraction zero?
- Think of $\frac{0}{5}$, $\frac{5}{5}$, and $\frac{5}{0}$: only the first is zero.
- A fraction is zero exactly when its top is zero — the bottom just cannot be zero too.
- Rational functions follow the same simple rule.
- So finding their zeros means looking only at the numerator.
Zeros come from the numerator
- A rational function 有理函数 is $\frac{\text{numerator}}{\text{denominator}}$.
- Its zeros 零点 are the inputs that make the whole thing equal $0$.
- That happens only where the numerator 分子 equals $0$.
- So set the top equal to zero and solve — those are your candidate x-intercepts.
A fraction equals zero exactly when…
A fraction is zero only when its top is zero (and its bottom is not). $\frac{0}{5} = 0$, but $\frac{5}{0}$ is undefined.
…but only if they are allowed
- A candidate zero counts only if the value is in the domain 定义域 of the function.
- If it also makes the denominator 分母 zero, the function is undefined there — no zero.
- So check each numerator zero against the bottom before you trust it.
Where does it cross, and where does it break?
y = a/(x − b) + c
The curve crosses the x-axis at its zero, but breaks at x = b where the denominator is zero. Those are two different places.
A numerator zero is a real zero of the function only if that value lies in the ____ of the function.
If a value makes both top and bottom zero, it is not in the domain, so it is a hole, not a zero.
Excluded values are not zeros
- A value that makes only the denominator zero is excluded from the domain.
- There the function is undefined — it produces a vertical asymptote or a hole, not a zero.
- Never call an excluded value an x-intercept.
- Zero on top → a zero of the function; zero on the bottom → trouble, not a zero.
A value that makes the denominator zero is a zero of the function.
A denominator zero is excluded from the domain — the function is undefined there, not zero.
Verify by evaluating
- To confirm a zero, substitute the value and check that the function really gives $0$.
- $f(x) = \frac{x-4}{x+1}$ at $x = 4$: $\frac{0}{5} = 0$. Yes — $x = 4$ is a zero.
- The same test rejects an excluded value, because you would be dividing by zero.
- Evaluating is the reliable final check.
How many zeros does $f(x) = \dfrac{(x-3)(x+2)}{x^2+1}$ have?
The numerator is zero at $x = 3$ and $x = -2$; the denominator $x^2+1$ is never zero, so both are valid — $2$ zeros.
Select all true statements about zeros of a rational function.
A denominator zero is excluded, not an x-intercept. The other three are correct.
Do not confuse a numerator zero (a real zero, an x-intercept) with a denominator zero (an excluded value). They are opposite: one is where the graph touches the axis, the other is where the graph breaks.
Find the zeros of $f(x) = \dfrac{(x-2)(x+3)}{x-2}$.
- Numerator zero candidates: $x = 2$ and $x = -3$.
- But $x = 2$ also makes the denominator zero, so it is excluded — a hole, not a zero.
- Only $x = -3$ is a real zero of the function.
The zeros of a rational function are the zeros of its numerator that lie in the domain. A value that makes the denominator zero is excluded, not a zero. Always check numerator zeros against the bottom, and verify by evaluating.