Rational Functions and Holes
Introduced| English |
|---|
| hole/həʊl/ |
| factor/ˈfæktə/ |
| numerator/ˈnjuːməreɪtə/ |
| denominator/dɪˈnɒmɪneɪtə/ |
| removable discontinuity/rɪˈmuːvəbl dɪskɒntɪˈnjuːɪti/ |
| domain/dəˈmeɪn/ |
A single missing point
- Most rational graphs break dramatically at vertical asymptotes.
- But sometimes the graph is a perfectly ordinary line or curve — with just one point punched out.
- That tiny gap is called a hole, and it is easy to miss.
- It appears for a very specific algebraic reason.
Where holes come from
- A hole 空心 appears when a factor 因式 cancels from both the numerator and the denominator.
- Take $\frac{(x-1)(x+2)}{(x-1)}$: the $(x-1)$ cancels, leaving $x+2$.
- The simplified graph is the smooth line $y = x + 2$.
- But at $x = 1$ the original fraction was $\frac{0}{0}$ — undefined — so a point is missing.
A hole in a rational graph appears where…
A common factor top and bottom cancels, so the simplified graph is smooth — but the original value is still excluded, leaving a hole.
Hole versus vertical asymptote
- Both come from a denominator 分母 zero, but they behave oppositely.
- Cancels with the numerator 分子 → a removable discontinuity 可去间断点: a hole.
- Does not cancel → a vertical asymptote where the graph shoots off.
- So always factor and simplify first, then see which zeros cancel.
The simplified function is just a line — with one point removed
y = ax + b
This is the simplified form of (x²−1)/(x−1). The graph is the line y = x + 1, but the real function has a hole at x = 1 where the original was undefined.
What distinguishes a hole from a vertical asymptote?
Cancelled factor → removable hole; surviving denominator zero → vertical asymptote where values blow up.
To find a hole's coordinates, cancel the common factor, then evaluate the ____ function at the excluded value.
The simplified function gives the y-value the graph would have there — that is the hole's height.
Finding the hole's coordinates
- Cancel the common factor to get the simplified function.
- Substitute the excluded x-value into that simplified function to get the y-value.
- For $\frac{(x-1)(x+2)}{x-1}$ at $x = 1$: simplified is $x+2$, giving $y = 3$.
- So the hole sits at $(1, 3)$ — the point the graph "should" have passed through.
Even after the factor cancels, the hole's x-value is still excluded from the domain.
The original function was never defined there, so cancelling in algebra does not add the point back — the domain still has a gap.
Select all true statements about a hole.
A hole does not send the function to infinity — that is an asymptote. The other three are true.
Still excluded from the domain
- Cancelling is algebra; it does not restore the missing input.
- The original function was undefined at that x, so it stays out of the domain 定义域.
- The simplified formula agrees with the original everywhere except that one point.
- That is exactly what "removable" means: you can patch the single gap, but it was still there.
Simplifying $\frac{(x-1)(x+2)}{x-1}$ to $x+2$ is only valid for $x \neq 1$. Never drop the restriction — the two expressions are equal everywhere except at the hole, and forgetting that loses a point from the domain.
Locate any hole of $f(x) = \dfrac{x^2 - 4}{x - 2}$.
- Factor the top: $\frac{(x-2)(x+2)}{x-2}$, and cancel $(x-2)$ → simplified $x + 2$.
- Excluded value $x = 2$; simplified value there is $2 + 2 = 4$.
- So there is a hole at $(2, 4)$, and no vertical asymptote.
A hole is a removable discontinuity: a factor that cancels from numerator and denominator. Find it by simplifying and evaluating at the excluded x-value. The x stays out of the domain — cancelling patches the graph but never adds the point back.