If a rational function is known to belong to the family f(x) = k/(x − h) + v, with k ≠ 0, its vertical asymptote is x = h and its horizontal asymptote is y = v. One additional point then determines the multiplier k. This is a function-reconstruction question rather than a request to guess a graph from its appearance.
For example, suppose the vertical asymptote is x = 2, the horizontal asymptote is y = 3, and the curve passes through (4,7). Write f(x) = 3 + k/(x − 2). Substitution gives 7 = 3 + k/2, so k = 8. The function is f(x) = 3 + 8/(x − 2), with x ≠ 2. This guide explains the assumptions needed for uniqueness, how to verify both asymptotes, and how to recover intercepts, inverse and graphical symmetry.
An asymptote is information about graph behaviour
A vertical asymptote x = h marks a line near which the magnitude of a rational function can become arbitrarily large. A horizontal asymptote y = v describes a level the function approaches as x moves far to the left or right.
In the special family f(x) = k/(x − h) + v, with k ≠ 0, those asymptotes are apparent directly from the denominator and vertical shift.
The horizontal shift h determines the excluded input, while v determines the level of the horizontal approach. The constant k controls scaling and which side of the horizontal asymptote each branch occupies.
Reconstructing a formula from these features requires selecting the right family first. The asymptotes alone do not uniquely determine an arbitrary rational function; the stated reciprocal form supplies the structural assumption that makes one more point sufficient.
Start with the correct reciprocal-function template
The basic reciprocal function is y = 1/x, with asymptotes x = 0 and y = 0.
A horizontal shift to x = h gives 1/(x − h). Multiplying by k stretches the outputs and may reflect the graph across its horizontal midline. Adding v shifts the whole graph vertically.
Thus the useful form is f(x) = v + k/(x − h). The vertical asymptote is x = h, and the horizontal asymptote is y = v when k ≠ 0.
This is not the general form of every rational function. It is a specific translated-and-scaled reciprocal family. Read the question’s stated or implied function form before adopting it.
Read the asymptotes in the worked example
Suppose the curve has vertical asymptote x = 2 and horizontal asymptote y = 3, and is known to be a translated reciprocal of the form v + k/(x − h).
The vertical information gives h = 2. The horizontal information gives v = 3.
Therefore f(x) = 3 + k/(x − 2). Only one unknown k remains.
Writing the template immediately is more useful than drawing several possible branches and trying to choose a scale visually. The asymptotes fix the position of the graph; the given coordinate will fix the amount and sign of its departure from y = 3.
Use the known point to recover k
The curve passes through (4,7), so f(4) = 7. Substitute into the partially reconstructed formula:
7 = 3 + k/(4 − 2) = 3 + k/2. Subtract 3 to obtain 4 = k/2, then multiply by 2 to get k = 8.
The completed function is f(x) = 3 + 8/(x − 2). The original domain excludes x = 2 because the denominator vanishes.
Check the point: f(4) = 3 + 8/2 = 7. The data now fit exactly, without any need for a graphing calculator or quadratic equation.
The one-line reconstruction formula
More generally, suppose the asymptotes are x = h and y = v, and the graph passes through P(p,q) where p ≠ h.
From q = v + k/(p − h), rearrange to k = (q − v)(p − h).
This formula expresses the scale as a product of the vertical displacement of P from the horizontal asymptote and the horizontal displacement from the vertical asymptote.
For h = 2, v = 3 and P(4,7), the displacements are 2 and 4, so k = 4·2 = 8.
A student need not memorise the formula in isolation. Writing and solving the single point-substitution equation reveals both its meaning and its necessary conditions.
Check both asymptotes of the recovered function
For f(x) = 3 + 8/(x − 2), the denominator approaches zero as x approaches 2, so the reciprocal term becomes unbounded in magnitude. This establishes the vertical asymptote x = 2.
As x approaches positive infinity or negative infinity, 8/(x − 2) tends to zero. Therefore f(x) approaches 3, establishing horizontal asymptote y = 3.
The asymptote verification is independent of substituting the known point. Matching one coordinate alone does not establish that the entire function has the requested graph behaviour.
The three conditions—one vertical asymptote, one horizontal asymptote and one ordinary point—are all satisfied by the reconstructed member of the specified reciprocal family.
The domain is every real number except the vertical asymptote
The denominator x − 2 is zero at x = 2. Therefore the function has natural real domain {x ∈ ℝ : x ≠ 2}.
The asymptote line x = 2 is not an allowed input. A function cannot assign a finite value there using the displayed rational formula.
This restriction remains important even if a later expression is rewritten algebraically. Domain is part of the function, not merely a decoration added after differentiation or graphing.
When writing the final answer, state the formula and original domain together, especially if a follow-up asks for an inverse or composition. The excluded input becomes an excluded output of the inverse.
The range excludes the horizontal asymptote
The recovered function cannot equal 3 at any allowed x, because f(x) = 3 would require 8/(x − 2) = 0.
A rational expression with constant nonzero numerator 8 cannot have value zero for any finite allowed denominator. Therefore the range is {y ∈ ℝ : y ≠ 3}.
This is more than a graphical guess. It follows directly from the algebra and the condition k ≠ 0.
The range restriction later becomes the inverse function’s input restriction. It also explains why a point whose y-coordinate equals the horizontal asymptote cannot belong to a genuine member of this reciprocal family.
Recover the x-intercept exactly
To find where the graph meets the x-axis, solve f(x) = 0.
Set 3 + 8/(x − 2) = 0. Then 8/(x − 2) = −3, giving 8 = −3(x − 2) = −3x + 6.
Thus −3x = 2 and x = −2/3. The x-intercept is (−2/3,0).
Check the denominator at that input: −2/3 − 2 = −8/3, which is nonzero. Then f(−2/3) = 3 + 8/(−8/3) = 3 − 3 = 0, confirming the result.
Recover the y-intercept without confusing it with the horizontal asymptote
The y-intercept is found by substituting x = 0, provided zero is in the domain. Here 0 ≠ 2, so it is allowed.
Calculate f(0) = 3 + 8/(0 − 2) = 3 − 4 = −1. Thus the graph crosses the y-axis at (0,−1).
This is different from the horizontal asymptote y = 3. An asymptote describes large-|x| behaviour, whereas a y-intercept is the finite value at input zero.
The intercept check is useful for sketching. On the left branch, where x < 2, the reciprocal term is negative, so the output lies below the horizontal asymptote. A value of −1 at x = 0 is consistent with that structure.
Convert to a single fractional linear expression
The reconstructed function can be combined over one denominator:
f(x) = 3 + 8/(x − 2) = [3(x − 2) + 8]/(x − 2).
Simplifying the numerator gives f(x) = (3x + 2)/(x − 2), for x ≠ 2.
The ratio of leading coefficients 3/1 gives the horizontal asymptote y = 3. The denominator root x = 2 remains an uncancelled vertical asymptote because the numerator at 2 is 8, not zero.
This alternate form connects reciprocal transformations with ordinary algebraic fractions and provides a second route to the intercepts.
The multiplier determines which branch is above or below
For k = 8 > 0, the reciprocal part k/(x − 2) is positive when x > 2 and negative when x < 2.
Therefore f(x) > 3 on the right of the vertical asymptote and f(x) < 3 on the left.
If k were negative instead, these relationships would reverse. The asymptote positions would stay the same, but the branches would occupy the opposite quadrants relative to their intersection (2,3).
The sign of k is therefore geometric information about branch placement, while its magnitude controls how far the graph lies from the horizontal asymptote at a given horizontal displacement.
A negative-multiplier example
Suppose another curve from the same family has asymptotes x = 2 and y = 3 but passes through (4,1).
Write f(x) = 3 + k/(x − 2). Substitute the point: 1 = 3 + k/2, so k/2 = −2 and k = −4.
Thus f(x) = 3 − 4/(x − 2). For x > 2 its outputs lie below 3, matching the given point (4,1). For x < 2 its outputs lie above 3.
This example shows why knowing only the asymptotes cannot select the correct branch orientation or scale. The extra point supplies that missing information.
Find the graph’s centre of symmetry
The vertical and horizontal asymptotes intersect at (h,v). For a translated reciprocal function, that intersection is the centre of 180° rotational symmetry.
In the worked example, the centre is (2,3). For any nonzero t, f(2 + t) = 3 + 8/t and f(2 − t) = 3 − 8/t.
Their outputs add to 6, so the graph points at horizontal inputs 2 + t and 2 − t have midpoint y-coordinate 3. Their x-coordinate midpoint is 2.
Thus a half-turn around (2,3) maps the graph onto itself. This supplies a strong graphical check and relates reciprocal function transformations to rotational symmetry discussed for cubic curves.
Use symmetry to find another point
The supplied point (4,7) lies two units right and four units above the centre (2,3).
Rotate it 180° about that centre. The partner point is two units left and four units below: (0,−1).
This is exactly the y-intercept found by direct substitution. The symmetry provides an independent check without repeating the fraction calculation.
More generally, if (p,q) is on the graph, the symmetric point is (2h − p, 2v − q). The formula works because both horizontal and vertical offsets change sign under a half-turn.
Differentiate to understand branch direction
For f(x) = 3 + 8/(x − 2), differentiate to obtain f′(x) = −8/(x − 2)².
The denominator square is positive for every allowed x, so the derivative is strictly negative on both separate domain intervals x < 2 and x > 2.
Therefore each branch is decreasing as x increases within that interval. This is consistent with the right branch descending from large positive values toward 3, and the left branch descending toward negative infinity as it approaches x = 2 from the left.
The function is not continuous across x = 2, so the two decreasing branches should not be described as one continuous decreasing path through the excluded input.
Find the inverse function from the reconstructed formula
Write y = 3 + 8/(x − 2). Subtract 3: y − 3 = 8/(x − 2).
Multiply and rearrange: (y − 3)(x − 2) = 8, so x − 2 = 8/(y − 3), and x = 2 + 8/(y − 3).
Swap labels to obtain f⁻¹(x) = 2 + 8/(x − 3), with domain x ≠ 3.
The original function has excluded input 2 and excluded output 3. The inverse exchanges those restrictions. Checking the composition f(f⁻¹(x)) = x on x ≠ 3 confirms the reversed mapping.
A point on the proposed vertical asymptote is impossible
Suppose a problem states that the curve f(x) = 3 + k/(x − 2) passes through (2,7).
This is impossible because x = 2 makes the original denominator zero for any nonzero k. The point is not in the function’s domain.
Do not attempt to determine k by multiplying a zero denominator through the equation 7 = 3 + k/0. Division by zero is not a legitimate operation.
A supplied point for this reconstruction must have p ≠ h. Checking that condition before calculating saves an invalid substitution and identifies contradictory data clearly.
A point on the proposed horizontal asymptote is also incompatible
Suppose the asymptotes are still x = 2 and y = 3, but the alleged point is (4,3).
The point substitution gives 3 = 3 + k/2, hence k = 0. Then the function would be constant 3 for x ≠ 2, not a genuine reciprocal with a vertical asymptote.
Therefore the stated conditions are incompatible with the requirement k ≠ 0. The asymptote data and point cannot all hold for a nondegenerate member of the selected family.
This is an information-checking lesson. An algebraic parameter value may solve one substituted equation but violate a structural assumption used to define the problem.
The asymptotes alone do not determine k
Suppose only x = 2 and y = 3 are supplied. The functions 3 + 8/(x − 2), 3 − 4/(x − 2) and 3 + 1/(x − 2) all share those asymptotes.
They differ in their positions relative to the asymptotes and in their values at ordinary input points. Therefore infinitely many curves in this family remain possible.
One allowed point that does not lie on either asymptote supplies the missing scale k and selects a unique member of the family.
If a question lacks that additional condition, the correct general answer is f(x) = 3 + k/(x − 2), with k any nonzero real constant. Do not invent an arbitrary numerical k simply to produce a single graph.
Even two asymptotes and one point do not determine every rational function
Uniqueness depends on assuming the reciprocal template. A more general rational function can share the same two asymptotes and point while having additional terms.
For example, let g(x) = 3 + 8/(x − 2) + (x − 4)/(x − 2)². At x = 4, the added term is zero, so g(4) = 7.
As x tends to large positive or negative values, both rational terms tend to zero, so y = 3 remains the horizontal asymptote. The denominator still produces vertical-asymptote behaviour near x = 2.
Yet g is not the same function as 3 + 8/(x − 2). Thus the model family must be stated or reasonably supplied by the task; the graph features alone do not uniquely reconstruct an unrestricted rational expression.
A slant-asymptote extension
A related family is f(x) = mx + c + k/(x − h), where the rational correction tends to zero as |x| grows.
Its vertical asymptote is x = h when k ≠ 0, while its slant asymptote is the line y = mx + c rather than a horizontal constant.
Suppose h = 2, the slant asymptote is y = x + 1, and the graph passes through (4,8). Then 8 = 4 + 1 + k/(4 − 2), giving k = 6.
The reconstructed member is f(x) = x + 1 + 6/(x − 2). This is an extension of the same displacement-and-scale reasoning, but it should not be confused with the simpler horizontal-asymptote template.
A second example with a negative vertical asymptote
Suppose a translated reciprocal has vertical asymptote x = −1, horizontal asymptote y = 2 and passes through (1,5).
Write f(x) = 2 + k/(x + 1). Substitute (1,5): 5 = 2 + k/2, giving k = 6.
Therefore f(x) = 2 + 6/(x + 1). Its domain excludes x = −1 and its range excludes y = 2.
Check the point: f(1) = 2 + 6/2 = 5. The method is identical to the first example, even though the denominator now has a plus sign and the asymptote lies left of the y-axis.
A quick reconstruction and verification routine
First verify that the question specifies the family f(x) = v + k/(x − h), with k ≠ 0. Read h from the vertical asymptote and v from the horizontal asymptote.
Second substitute the supplied point (p,q) and solve k = (q − v)(p − h). Check p ≠ h and k ≠ 0.
Third write the full formula and original domain. Verify the point, the two asymptotes and the predicted branch position.
For an extended question, find intercepts, range, inverse or rotational symmetry as requested. This short sequence exposes the structural assumptions and stops a neat-looking formula from being accepted without valid domain and range checks.
Practice tasks with checked answers
Question one: f(x) = v + k/(x − h) has asymptotes x = 2 and y = 3, and passes through (4,7). Answer: f(x) = 3 + 8/(x − 2).
Question two: same asymptotes but point (4,1). Answer: f(x) = 3 − 4/(x − 2).
Question three: asymptotes x = −1 and y = 2, through (1,5). Answer: f(x) = 2 + 6/(x + 1).
Question four: for Question one, find the x-intercept and y-intercept. Answer: (−2/3,0) and (0,−1).
Question five: for Question one, find the inverse. Answer: f⁻¹(x) = 2 + 8/(x − 3), with x ≠ 3.
Question six: can a genuine reciprocal with asymptotes x = 2 and y = 3 pass through (4,3)? No. That would force k = 0 and remove the vertical asymptote.
For each, check the given point and both asymptotes before concluding.
Syllabus scope and related learning
Reciprocal-function reconstruction connects rational graph transformations, asymptotes, domain and range, inverse functions and coordinate substitution. Some reconstruction and slant-asymptote extensions may be enrichment according to the student’s course.
Singapore families should consult the appropriate SEAB syllabus lists for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2.
Related guides include How to Invert a Fractional Linear Function Without Losing Its Domain, How to Find a Slant Asymptote by Polynomial Division and How to Distinguish a Hole from a Vertical Asymptote in a Rational Graph. Browse the Additional Mathematics archive or enquire about suitable eduKate Singapore small-group tutorials near Sixth Avenue MRT.
The enduring skill is knowing exactly what a family of functions assumes, which graph features determine its position and how one independent point fixes its remaining scale. Properly taught kids shine a bright light into the future.