JC2 H2 Maths Bukit Timah | Hole vs Asymptote

Did you know? Cancelling (x−2) in (x²−x−2)/(x−2) gives x+1, but it does not put the missing input x=2 back into the original function. The graph follows the line y=x+1 everywhere except one missing point. That point is a hole, not a vertical asymptote. In JC2 H2 Mathematics, the small domain restriction you carry through your algebra decides which graph actually represents the function.

This JC2 H2 Mathematics Tutor Bukit Timah guide answers a narrow A-Level 9758 question: when does a denominator zero produce a removable hole, and when does it produce a vertical asymptote? We compare two functions with nearly identical numerators, show the cancellation rule without losing the domain, and provide worked graph checks that a student can reproduce with or without a graphing calculator. The quick answer is at the top because the first decision takes only seconds once understood.

For JC2 H2 Mathematics learners in Bukit Timah, this distinction connects rational graphs, algebraic cancellation, domain restrictions and asymptotes. The 2027 Singapore–Cambridge H2 Mathematics syllabus 9758 includes domain and range, rational graphs of specified types, asymptotes and restrictions on possible input values. This article builds directly on that foundation. All examples are original mathematical constructions, not claims about actual students or official examination questions.

Quick answer: a common factor removed leaves a hole; an uncancelled zero can create a vertical asymptote

Start with the original fraction, not just the simplified expression. If the denominator vanishes at x=a, then a is excluded from the original domain. If a matching numerator factor also vanishes and cancels, the simplified expression may approach a finite number as x→a, but the original function remains undefined at a: the graph has a removable hole. If the denominator still tends to zero while the numerator tends to a nonzero number, the function grows without bound in magnitude: there is a vertical asymptote at that input.

Before cancelling: solve the denominator =0 and write the excluded x-values.
After cancelling: simplify the algebra but retain the exclusions from the original expression.
For a cancelled zero: evaluate the simplified expression there to locate the missing point on its graph.
For an uncancelled zero: check that the remaining numerator is nonzero; then investigate one-sided behaviour.
For any potential asymptote: remember that a denominator zero by itself is not sufficient proof.
For calculator plotting: a thin pixel-width gap may be invisible; use algebra to distinguish a hole from a vertical line.

First worked comparison: one numerator, two denominators

Compare f(x)=[(x−2)(x+1)]/(x−2) with g(x)=[(x−2)(x+1)]/(x−3). The numerator is identical; only the denominator changes. The expression for f simplifies to x+1 whenever x≠2. Since its original denominator is zero at x=2, the graph of f is the line y=x+1 with the point (2,3) removed. The limiting value near x=2 is 3, but the function value f(2) itself is undefined.

The denominator of g vanishes at x=3. Its numerator there is (3−2)(3+1)=4, nonzero, so the fraction has a vertical asymptote at x=3. The sign of the numerator near 3 is positive: as x approaches 3 from the right, g(x) tends to +∞; from the left, it tends to −∞. There is no cancelled hole at x=2 in this second function because its denominator does not vanish at x=2; in fact g(2)=0.

Notice how three distinct statements appear: f is undefined at x=2 but has a finite limiting value there; g is defined and zero at x=2; g is undefined with unbounded behaviour near x=3. The central lesson is that the original denominator and residual numerator determine the graph. Two rational expressions can look closely related while representing different domain restrictions and vertical behaviour.

Why cancelling a factor does not restore a forbidden input

Algebraically, (x−a)/(x−a)=1 is valid only when x−a is nonzero. At x=a it is the undefined form 0/0, not the number one. Therefore simplifying (x−2)(x+1)/(x−2) to x+1 is an equality on the common domain x≠2, not an assertion that the original formula is defined at x=2. The deleted factor leaves a trace in the domain even when it disappears from the simplified expression.

A function is more than its visible arithmetic rule. Its domain matters. For the given f, the precise description is f(x)=x+1 for all real x except 2. If someone defines a new function h(x)=x+1 for every real x, h and f agree wherever both are defined, but h has an extra input and is therefore not the same function on the same domain. The missing point is precisely what the graph should indicate.

How to calculate the coordinates of a hole

To find a removable hole, retain the original exclusion x=a, cancel the common factor and substitute a into the simplified expression to obtain the limiting y-coordinate. In f(x)=(x−2)(x+1)/(x−2), the original function excludes x=2. The simplified rule is y=x+1, so the missing point is at (2,3). Mark a small open circle there, not a filled dot. A filled dot would falsely claim f(2)=3.

If the graphing calculator traces straight through the point at (2,3), that may reflect the device’s sampling and rendering resolution. It is not evidence that the original function has acquired a value. A calculator is useful for a general shape, but the exact domain must come from algebra. This is especially important in a test where the graph appears indistinguishable from an uninterrupted line on a small screen.

A vertical asymptote is about behaviour near the excluded input

For g(x)=(x−2)(x+1)/(x−3), the denominator approaches zero as x approaches 3. Meanwhile, the numerator approaches 4, not zero. Thus the magnitude of the quotient becomes arbitrarily large near the exclusion. The line x=3 is a vertical asymptote; it is not part of the graph. If the numerator limit were negative, the +∞ and −∞ one-sided directions would reverse, but the asymptote would remain at the same x-value.

Use a sign chart rather than guessing the direction. For x just above 3, x−3 is positive; for x just below 3, it is negative. The numerator is positive around this point, so the ratio is positive large on the right and negative large on the left. This is a stronger explanation than simply saying “denominator zero means asymptote”, which is false in a cancelled-factor case.

A second graph: an oblique asymptote is not a hole

Polynomial division also reveals g’s end behaviour. Expanding the numerator gives x²−x−2. Dividing by x−3 gives g(x)=x+2+4/(x−3), for x≠3. The rational correction 4/(x−3) approaches zero as x grows without bound in magnitude, so the graph approaches the oblique asymptote y=x+2 far from the vertical line. This is distinct from the vertical asymptote x=3 and from the isolated hole in f.

The distinction is geometric: a vertical asymptote describes unbounded behaviour near a finite forbidden input; an oblique asymptote describes a linear trend as x moves far left or far right. A hole describes a single omitted point where the limit is finite. H2 questions on rational graphs can test more than one of these characteristics at once, so name each using the behaviour it describes.

When repeated numerator and denominator factors partially cancel

Consider (x−1)²/(x−1), with x≠1. Cancelling one copy gives x−1 on its inherited domain. The limiting value at x=1 is zero, so the graph has a hole at (1,0). By contrast, (x−1)/(x−1)² simplifies to 1/(x−1) for x≠1 and has a vertical asymptote. In both examples a factor cancels, but the power of that factor still left in the denominator differs.

This is why you must simplify completely and inspect the resulting denominator, not merely check whether “some factor cancels”. In the second example one denominator factor survives, so the quotient is unbounded near x=1. In the first example no denominator factor survives, so its limiting value is finite. The multiplicities provide the correct decision.

Removable discontinuity is not the same as an x-intercept

In f(x)=(x−2)(x+1)/(x−2), the visible graph y=x+1 crosses the x-axis at x=−1. That point belongs to the original domain, so f(−1)=0 and it is a genuine x-intercept. The cancelled numerator zero at x=2, however, cannot be claimed as an x-intercept because f(2) is undefined. The algebraic root x=2 was present in the numerator, but it did not survive as an actual zero of the function.

For g(x)=(x−2)(x+1)/(x−3), both numerator roots x=2 and x=−1 remain in its domain, so both are genuine x-intercepts. The excluded input x=3 is not an intercept; it is a vertical asymptote. A student should state “zeros of the numerator that do not also make the denominator zero”, rather than indiscriminately listing every numerator factor as a graph crossing.

What happens to the range when you remove a point?

Take f(x)=(x−2)(x+1)/(x−2), with x≠2. Its simplified rule y=x+1 is a one-to-one line. The missing input x=2 would have produced output 3, and no other input produces y=3. Therefore the range of f is all real y except 3. This is an especially valuable consequence of preserving the domain: a removable hole can remove a value from the range as well as the domain.

That conclusion is not automatic for every function. If a horizontal or curved graph takes the same y-value elsewhere, deleting one input may leave the corresponding y-value in the range. For example, a quadratic with a missing point could still reach that output at another input. State the range using the actual simplified rule and remaining domain, not just the words “one point is missing”.

Three questions to ask before using a calculator graph

Which inputs are forbidden? List every original denominator zero, even if factors cancel later.
What finite limit or unbounded pattern remains? Determine which exclusions are holes and which are vertical asymptotes.
What does the simplified rule predict elsewhere? Use it for intercepts and end behaviour while respecting the exclusions.

A graphing calculator is particularly valuable for estimating the location of a visible vertical asymptote and viewing a graph’s overall shape, but many screens do not display a removable hole as an obvious open circle. For algebraic justification, compute the hole coordinate exactly and label it. The 2027 H2 Mathematics syllabus expects candidates to use an approved graphing calculator, while also recognising its limitations and providing mathematical working when required.

One-minute Bukit Timah diagnostic

Show a learner (x−2)(x+1)/(x−2). Ask first for the forbidden x-value, then the simplified expression, then the graph feature. A correct sequence is x≠2, y=x+1, hole at (2,3). Without changing the numerator, replace the denominator by x−3. Now x=3 is excluded, the numerator at that input is 4, and x=3 becomes a vertical asymptote. This changed-case method isolates the role of the denominator.

For a final check, ask whether x=2 is an x-intercept for the first function. It is not, because the function is undefined there. Ask whether x=2 is an x-intercept for the second function. It is, because the denominator x−3 remains nonzero there. A learner who can justify both answers from the domain has learned a transferable concept rather than a slogan about cancelling factors.

Thirty-six original hole-versus-asymptote clinics

Each exercise below uses two rational functions sharing a quadratic numerator but having a different linear denominator. Their forms belong to the kind of rational-graph reasoning used in H2 functions and graphs. The integers a, b and c are deliberately distinct to make the comparison checkable. All functions are fictional mathematics examples. In each clinic, one denominator cancels to make a hole; the other does not and produces a vertical asymptote.

1. Hole at x=-6, asymptote at x=2

Start with an original pair of functions. Let f(x)=(x+6)(x+3)/(x+6), and let g(x)=(x+6)(x+3)/(x−2). All three constants -6, -3 and 2 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-6, but the matching factor cancels for other inputs. Its simplified expression is y=x + 3 for x≠-6. The missing coordinate is (-6, -3), because the simplified rule approaches -3 at the forbidden input. Its genuine x-intercept is (-3,0); the cancelled root x=-6 is not in the domain. Since the simplified rule is one-to-one, output y=-3 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-6)(2−-3)=40, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 11 far from this x-value. Do not fill the open circle merely because the simplified line has that coordinate. The original function was undefined there. The pair’s difference comes from whether a denominator factor remains after simplification.

2. Hole at x=-3, asymptote at x=-6

Start with an original pair of functions. Let f(x)=(x+3)(x−3)/(x+3), and let g(x)=(x+3)(x−3)/(x+6). All three constants -3, 3 and -6 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-3, but the matching factor cancels for other inputs. Its simplified expression is y=x − 3 for x≠-3. The missing coordinate is (-3, -6), because the simplified rule approaches -6 at the forbidden input. Its genuine x-intercept is (3,0); the cancelled root x=-3 is not in the domain. Since the simplified rule is one-to-one, output y=-6 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-3)(-6−3)=27, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 6 far from this x-value. For the second fraction, the denominator zero survives, and the numerator is nonzero. That confirms a vertical asymptote. The pair’s difference comes from whether a denominator factor remains after simplification.

3. Hole at x=1, asymptote at x=2

Start with an original pair of functions. Let f(x)=(x−1)(x+7)/(x−1), and let g(x)=(x−1)(x+7)/(x−2). All three constants 1, -7 and 2 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=1, but the matching factor cancels for other inputs. Its simplified expression is y=x + 7 for x≠1. The missing coordinate is (1, 8), because the simplified rule approaches 8 at the forbidden input. Its genuine x-intercept is (-7,0); the cancelled root x=1 is not in the domain. Since the simplified rule is one-to-one, output y=8 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−1)(2−-7)=9, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 8 far from this x-value. A simplified function and the original fractional expression can agree on every permitted input yet have different stated domains. The pair’s difference comes from whether a denominator factor remains after simplification.

4. Hole at x=4, asymptote at x=-6

Start with an original pair of functions. Let f(x)=(x−4)(x+2)/(x−4), and let g(x)=(x−4)(x+2)/(x+6). All three constants 4, -2 and -6 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=4, but the matching factor cancels for other inputs. Its simplified expression is y=x + 2 for x≠4. The missing coordinate is (4, 6), because the simplified rule approaches 6 at the forbidden input. Its genuine x-intercept is (-2,0); the cancelled root x=4 is not in the domain. Since the simplified rule is one-to-one, output y=6 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−4)(-6−-2)=40, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 8 far from this x-value. Look for genuine x-intercepts in the original domain. A cancelled numerator root at the forbidden input is not a valid zero. The pair’s difference comes from whether a denominator factor remains after simplification.

5. Hole at x=7, asymptote at x=2

Start with an original pair of functions. Let f(x)=(x−7)(x−4)/(x−7), and let g(x)=(x−7)(x−4)/(x−2). All three constants 7, 4 and 2 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=7, but the matching factor cancels for other inputs. Its simplified expression is y=x − 4 for x≠7. The missing coordinate is (7, 3), because the simplified rule approaches 3 at the forbidden input. Its genuine x-intercept is (4,0); the cancelled root x=7 is not in the domain. Since the simplified rule is one-to-one, output y=3 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−7)(2−4)=10, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 9 far from this x-value. The vertical asymptote tells you about behaviour close to a finite excluded x, not behaviour far from the graph. The pair’s difference comes from whether a denominator factor remains after simplification.

6. Hole at x=-5, asymptote at x=-7

Start with an original pair of functions. Let f(x)=(x+5)(x+6)/(x+5), and let g(x)=(x+5)(x+6)/(x+7). All three constants -5, -6 and -7 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-5, but the matching factor cancels for other inputs. Its simplified expression is y=x + 6 for x≠-5. The missing coordinate is (-5, 1), because the simplified rule approaches 1 at the forbidden input. Its genuine x-intercept is (-6,0); the cancelled root x=-5 is not in the domain. Since the simplified rule is one-to-one, output y=1 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-7. The numerator approaches (-7−-5)(-7−-6)=2, which is nonzero, so the vertical asymptote is x=-7. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 4 far from this x-value. A hole can be nearly invisible on a screen. Determine its exact coordinate algebraically. The pair’s difference comes from whether a denominator factor remains after simplification.

7. Hole at x=-2, asymptote at x=2

Start with an original pair of functions. Let f(x)=(x+2)(x+1)/(x+2), and let g(x)=(x+2)(x+1)/(x−2). All three constants -2, -1 and 2 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-2, but the matching factor cancels for other inputs. Its simplified expression is y=x + 1 for x≠-2. The missing coordinate is (-2, -1), because the simplified rule approaches -1 at the forbidden input. Its genuine x-intercept is (-1,0); the cancelled root x=-2 is not in the domain. Since the simplified rule is one-to-one, output y=-1 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-2)(2−-1)=12, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 5 far from this x-value. Changing the denominator without changing the numerator can completely alter graph topology. The pair’s difference comes from whether a denominator factor remains after simplification.

8. Hole at x=2, asymptote at x=-6

Start with an original pair of functions. Let f(x)=(x−2)(x−5)/(x−2), and let g(x)=(x−2)(x−5)/(x+6). All three constants 2, 5 and -6 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=2, but the matching factor cancels for other inputs. Its simplified expression is y=x − 5 for x≠2. The missing coordinate is (2, -3), because the simplified rule approaches -3 at the forbidden input. Its genuine x-intercept is (5,0); the cancelled root x=2 is not in the domain. Since the simplified rule is one-to-one, output y=-3 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−2)(-6−5)=88, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 13 far from this x-value. The sign of the residual numerator near the excluded x controls which side tends to positive infinity. The pair’s difference comes from whether a denominator factor remains after simplification.

9. Hole at x=5, asymptote at x=2

Start with an original pair of functions. Let f(x)=(x−5)(x+5)/(x−5), and let g(x)=(x−5)(x+5)/(x−2). All three constants 5, -5 and 2 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=5, but the matching factor cancels for other inputs. Its simplified expression is y=x + 5 for x≠5. The missing coordinate is (5, 10), because the simplified rule approaches 10 at the forbidden input. Its genuine x-intercept is (-5,0); the cancelled root x=5 is not in the domain. Since the simplified rule is one-to-one, output y=10 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−5)(2−-5)=-21, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to −∞, and on the left to +∞. Polynomial division gives an oblique asymptote y=x + 2 far from this x-value. A linear simplified rule can be extended continuously, but that extension is a new function if the original domain excluded the point. The pair’s difference comes from whether a denominator factor remains after simplification.

10. Hole at x=-7, asymptote at x=-6

Start with an original pair of functions. Let f(x)=(x+7)(x−1)/(x+7), and let g(x)=(x+7)(x−1)/(x+6). All three constants -7, 1 and -6 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-7, but the matching factor cancels for other inputs. Its simplified expression is y=x − 1 for x≠-7. The missing coordinate is (-7, -8), because the simplified rule approaches -8 at the forbidden input. Its genuine x-intercept is (1,0); the cancelled root x=-7 is not in the domain. Since the simplified rule is one-to-one, output y=-8 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-7)(-6−1)=-7, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to −∞, and on the left to +∞. Polynomial division gives an oblique asymptote y=x + 0 far from this x-value. Check the line’s slope and compare it with the oblique asymptote of the uncancelled rational expression. The pair’s difference comes from whether a denominator factor remains after simplification.

11. Hole at x=-4, asymptote at x=2

Start with an original pair of functions. Let f(x)=(x+4)(x−6)/(x+4), and let g(x)=(x+4)(x−6)/(x−2). All three constants -4, 6 and 2 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-4, but the matching factor cancels for other inputs. Its simplified expression is y=x − 6 for x≠-4. The missing coordinate is (-4, -10), because the simplified rule approaches -10 at the forbidden input. Its genuine x-intercept is (6,0); the cancelled root x=-4 is not in the domain. Since the simplified rule is one-to-one, output y=-10 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-4)(2−6)=-24, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to −∞, and on the left to +∞. Polynomial division gives an oblique asymptote y=x + 0 far from this x-value. Do not fill the open circle merely because the simplified line has that coordinate. The original function was undefined there. The pair’s difference comes from whether a denominator factor remains after simplification.

12. Hole at x=-1, asymptote at x=-6

Start with an original pair of functions. Let f(x)=(x+1)(x+4)/(x+1), and let g(x)=(x+1)(x+4)/(x+6). All three constants -1, -4 and -6 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=-1, but the matching factor cancels for other inputs. Its simplified expression is y=x + 4 for x≠-1. The missing coordinate is (-1, 3), because the simplified rule approaches 3 at the forbidden input. Its genuine x-intercept is (-4,0); the cancelled root x=-1 is not in the domain. Since the simplified rule is one-to-one, output y=3 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-1)(-6−-4)=10, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 1 far from this x-value. For the second fraction, the denominator zero survives, and the numerator is nonzero. That confirms a vertical asymptote. The pair’s difference comes from whether a denominator factor remains after simplification.

13. Hole at x=3, asymptote at x=1

Start with an original pair of functions. Let f(x)=(x−3)(x−2)/(x−3), and let g(x)=(x−3)(x−2)/(x−1). All three constants 3, 2 and 1 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=3, but the matching factor cancels for other inputs. Its simplified expression is y=x − 2 for x≠3. The missing coordinate is (3, 1), because the simplified rule approaches 1 at the forbidden input. Its genuine x-intercept is (2,0); the cancelled root x=3 is not in the domain. Since the simplified rule is one-to-one, output y=1 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=1. The numerator approaches (1−3)(1−2)=2, which is nonzero, so the vertical asymptote is x=1. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 4 far from this x-value. A simplified function and the original fractional expression can agree on every permitted input yet have different stated domains. The pair’s difference comes from whether a denominator factor remains after simplification.

14. Hole at x=6, asymptote at x=-6

Start with an original pair of functions. Let f(x)=(x−6)(x−7)/(x−6), and let g(x)=(x−6)(x−7)/(x+6). All three constants 6, 7 and -6 are distinct. Without using a calculator, list each original denominator exclusion, decide whether it becomes a removable hole or a vertical asymptote, and label every genuine x-intercept. Two functions that share a numerator need not share a graph shape.

Correct hole calculation. For f, the original domain excludes x=6, but the matching factor cancels for other inputs. Its simplified expression is y=x − 7 for x≠6. The missing coordinate is (6, -1), because the simplified rule approaches -1 at the forbidden input. Its genuine x-intercept is (7,0); the cancelled root x=6 is not in the domain. Since the simplified rule is one-to-one, output y=-1 is also absent from its range.

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−6)(-6−7)=156, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 19 far from this x-value. Look for genuine x-intercepts in the original domain. A cancelled numerator root at the forbidden input is not a valid zero. The pair’s difference comes from whether a denominator factor remains after simplification.

15. Hole at x=-6, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-6)(2−-3)=40, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 11 far from this x-value. The vertical asymptote tells you about behaviour close to a finite excluded x, not behaviour far from the graph. The pair’s difference comes from whether a denominator factor remains after simplification.

16. Hole at x=-3, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-3)(-6−3)=27, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 6 far from this x-value. A hole can be nearly invisible on a screen. Determine its exact coordinate algebraically. The pair’s difference comes from whether a denominator factor remains after simplification.

17. Hole at x=1, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−1)(2−-7)=9, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 8 far from this x-value. Changing the denominator without changing the numerator can completely alter graph topology. The pair’s difference comes from whether a denominator factor remains after simplification.

18. Hole at x=4, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−4)(-6−-2)=40, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 8 far from this x-value. The sign of the residual numerator near the excluded x controls which side tends to positive infinity. The pair’s difference comes from whether a denominator factor remains after simplification.

19. Hole at x=7, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−7)(2−4)=10, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 9 far from this x-value. A linear simplified rule can be extended continuously, but that extension is a new function if the original domain excluded the point. The pair’s difference comes from whether a denominator factor remains after simplification.

20. Hole at x=-5, asymptote at x=-7

Correct asymptote calculation. For g, the forbidden input is x=-7. The numerator approaches (-7−-5)(-7−-6)=2, which is nonzero, so the vertical asymptote is x=-7. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 4 far from this x-value. Check the line’s slope and compare it with the oblique asymptote of the uncancelled rational expression. The pair’s difference comes from whether a denominator factor remains after simplification.

21. Hole at x=-2, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-2)(2−-1)=12, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 5 far from this x-value. Do not fill the open circle merely because the simplified line has that coordinate. The original function was undefined there. The pair’s difference comes from whether a denominator factor remains after simplification.

22. Hole at x=2, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−2)(-6−5)=88, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 13 far from this x-value. For the second fraction, the denominator zero survives, and the numerator is nonzero. That confirms a vertical asymptote. The pair’s difference comes from whether a denominator factor remains after simplification.

23. Hole at x=5, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−5)(2−-5)=-21, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to −∞, and on the left to +∞. Polynomial division gives an oblique asymptote y=x + 2 far from this x-value. A simplified function and the original fractional expression can agree on every permitted input yet have different stated domains. The pair’s difference comes from whether a denominator factor remains after simplification.

24. Hole at x=-7, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-7)(-6−1)=-7, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to −∞, and on the left to +∞. Polynomial division gives an oblique asymptote y=x + 0 far from this x-value. Look for genuine x-intercepts in the original domain. A cancelled numerator root at the forbidden input is not a valid zero. The pair’s difference comes from whether a denominator factor remains after simplification.

25. Hole at x=-4, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-4)(2−6)=-24, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to −∞, and on the left to +∞. Polynomial division gives an oblique asymptote y=x + 0 far from this x-value. The vertical asymptote tells you about behaviour close to a finite excluded x, not behaviour far from the graph. The pair’s difference comes from whether a denominator factor remains after simplification.

26. Hole at x=-1, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-1)(-6−-4)=10, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 1 far from this x-value. A hole can be nearly invisible on a screen. Determine its exact coordinate algebraically. The pair’s difference comes from whether a denominator factor remains after simplification.

27. Hole at x=3, asymptote at x=1

Correct asymptote calculation. For g, the forbidden input is x=1. The numerator approaches (1−3)(1−2)=2, which is nonzero, so the vertical asymptote is x=1. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 4 far from this x-value. Changing the denominator without changing the numerator can completely alter graph topology. The pair’s difference comes from whether a denominator factor remains after simplification.

28. Hole at x=6, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−6)(-6−7)=156, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 19 far from this x-value. The sign of the residual numerator near the excluded x controls which side tends to positive infinity. The pair’s difference comes from whether a denominator factor remains after simplification.

29. Hole at x=-6, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-6)(2−-3)=40, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 11 far from this x-value. A linear simplified rule can be extended continuously, but that extension is a new function if the original domain excluded the point. The pair’s difference comes from whether a denominator factor remains after simplification.

30. Hole at x=-3, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−-3)(-6−3)=27, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 6 far from this x-value. Check the line’s slope and compare it with the oblique asymptote of the uncancelled rational expression. The pair’s difference comes from whether a denominator factor remains after simplification.

31. Hole at x=1, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−1)(2−-7)=9, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 8 far from this x-value. Do not fill the open circle merely because the simplified line has that coordinate. The original function was undefined there. The pair’s difference comes from whether a denominator factor remains after simplification.

32. Hole at x=4, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−4)(-6−-2)=40, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 8 far from this x-value. For the second fraction, the denominator zero survives, and the numerator is nonzero. That confirms a vertical asymptote. The pair’s difference comes from whether a denominator factor remains after simplification.

33. Hole at x=7, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−7)(2−4)=10, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 9 far from this x-value. A simplified function and the original fractional expression can agree on every permitted input yet have different stated domains. The pair’s difference comes from whether a denominator factor remains after simplification.

34. Hole at x=-5, asymptote at x=-7

Correct asymptote calculation. For g, the forbidden input is x=-7. The numerator approaches (-7−-5)(-7−-6)=2, which is nonzero, so the vertical asymptote is x=-7. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 4 far from this x-value. Look for genuine x-intercepts in the original domain. A cancelled numerator root at the forbidden input is not a valid zero. The pair’s difference comes from whether a denominator factor remains after simplification.

35. Hole at x=-2, asymptote at x=2

Correct asymptote calculation. For g, the forbidden input is x=2. The numerator approaches (2−-2)(2−-1)=12, which is nonzero, so the vertical asymptote is x=2. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x + 5 far from this x-value. The vertical asymptote tells you about behaviour close to a finite excluded x, not behaviour far from the graph. The pair’s difference comes from whether a denominator factor remains after simplification.

36. Hole at x=2, asymptote at x=-6

Correct asymptote calculation. For g, the forbidden input is x=-6. The numerator approaches (-6−2)(-6−5)=88, which is nonzero, so the vertical asymptote is x=-6. On the right, g tends to +∞, and on the left to −∞. Polynomial division gives an oblique asymptote y=x − 13 far from this x-value. A hole can be nearly invisible on a screen. Determine its exact coordinate algebraically. The pair’s difference comes from whether a denominator factor remains after simplification.

Ten short checks with answers

1. (x²−1)/(x−1), x≠1: y=x+1 with a hole at (1,2).
2. (x²−1)/(x−2): vertical asymptote at x=2, numerator there equals 3.
3. (x−1)²/(x−1): hole at (1,0).
4. (x−1)/(x−1)²: vertical asymptote at x=1.
5. (x−2)(x+1)/(x−2): x=2 is not an x-intercept.
6. The same function has genuine x-intercept at x=−1.
7. The range of x+1 restricted to x≠2 excludes y=3.
8. (x−2)(x+1)/(x−3) has asymptote x=3 and an x-intercept at x=2.
9. The latter rational graph has far-field oblique asymptote y=x+2.
10. Graphing through a hole does not make the original fractional function defined there.

A five-day problem-repair plan

Day 1: practise finding denominator zeros before cancelling any factors. Day 2: simplify cancelled expressions and label open-circle coordinates, explaining why the omitted input remains excluded. Day 3: handle uncancelled denominator zeros with a sign chart to predict one-sided infinity. Day 4: compare x-intercepts and domain restrictions across near-identical rational functions. Day 5: mix all three: hole, asymptote, and a rational function with no real denominator zero. The point is to form the habit of auditing the original domain independently of the simplified expression.

In an eduKateSG-style tutorial with up to three learners, one can be responsible for algebraic factoring, another for domain and range, and a third for graph features and a calculator audit. Rotate the roles and change just one factor. The learner who can explain why a hole becomes an asymptote when that factor moves out of the denominator has understood the mechanism more deeply than one who simply recognises a familiar plot.

Frequently asked questions

Is a denominator zero always a vertical asymptote?

No. A matching numerator factor may cancel, leaving a finite limit and a removable hole. Check the simplified denominator and numerator behaviour near the original exclusion.

If a factor cancels, can I include the excluded input?

Not in the original function. The simplification is valid only at inputs where the denominator was nonzero. A new continuous extension could define that point, but it would have a different domain.

How do I plot a removable hole?

Cancel the common factor, substitute the excluded x into the simplified expression to locate the finite limiting y, and mark an open circle at that coordinate.

Can a rational graph have an oblique and a vertical asymptote?

Yes. A quadratic numerator over a linear denominator can have a vertical asymptote at the denominator zero and an oblique trend from polynomial division.

What happens to the x-intercepts after cancellation?

A numerator root at a forbidden input is not an actual function zero. Only roots in the original domain can be genuine intercepts.

Will a graphing calculator show the hole?

Not reliably. It may draw an apparently continuous line over an isolated excluded input. Algebra is required for an exact domain statement.

Is this within the H2 Mathematics 9758 graph topic?

It reinforces the syllabus work on domain and range, rational graphs, asymptotes and input restrictions. The examples are explanatory extensions of familiar algebra, not exact copies of past-year questions.

What is the difference between a hole and a vertical asymptote?

At a removable hole the function is undefined at one input but approaches a finite number there. At a vertical asymptote, the magnitude becomes unbounded on approach from at least one side.

JC2 H2 Mathematics Tutor Bukit Timah | A Hole or a Vertical Asymptote?