01 · Signs
y = −6/x
Hint
A negative numerator has the opposite sign to x.
Worked solution
Positive branch for x < 0; negative branch for x > 0. Points include (−2, 3) and (2, −3). Asymptotes: x = 0, y = 0. Domain x ≠ 0; range y ≠ 0.
Understand · explore · practise
Reciprocal graphs
Understand y = k/x and y = k/x², sketch their branches and asymptotes, compare scales and find translated reciprocal graphs.
Before you startFractions, negative numbers and coordinate axes
01 / The two families
A squared denominator changes the left branch.
For a non-zero constant k, the functions k/x and k/x² are undefined at x = 0. Their graphs have two separate branches.
With k > 0, k/x is positive for positive x and negative for negative x. But k/x² is positive on both sides, because x² > 0 whenever x ≠ 0.
Changing k to a negative number reverses every output. Try both denominators and both signs in the graph.
k/x: opposite signs on opposite sides
k/x²: same sign on both sides
Domain: x ≠ 0
y = 3/xx ≠ 0
y = 3/x: positive on the right and negative on the left. At x = 1, y = 3; at x = −1, y = −3. Asymptotes: x = 0 and y = 0.
Watch the two reciprocal families
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Asymptotes
Describe what happens near a gap and far away.
A vertical asymptote describes unbounded outputs as x approaches a particular value from at least one side. A horizontal asymptote describes the value approached as x goes far to the left or right.
For k/x and k/x², with k ≠ 0:
vertical asymptote x = 0
horizontal asymptote y = 0
For 3/x, approaching zero from the right gives large positive outputs; from the left gives large negative outputs. For 3/x², both sides give large positive outputs. As |x| becomes large, both expressions approach zero.
These particular graphs never meet either axis: x = 0 is excluded, and a non-zero numerator cannot produce y = 0.
Two qualifications
“An asymptote can never be crossed” is not a general rule: other functions can cross a horizontal asymptote. Also, k = 0 is a separate case. Then 0/x is zero for x ≠ 0 and does not have a vertical asymptote at zero.
03 / Compare scales
Compare outputs at the same input.
Increasing a positive numerator from 2 to 6 triples each output. It moves each point vertically away from the x-axis, but leaves both asymptotes fixed.
x = 2: 2/x = 1, 6/x = 3
x = −2: 2/x = −1, 6/x = −3
So 6/x is above 2/x on the right, but below it on the left. For the squared denominators, 6/x² is above 2/x² on both sides.
The graph of k/x has rotational symmetry through the origin: f(−x) = −f(x). The graph of k/x² has symmetry in the y-axis: f(−x) = f(x).
Compare negative numeratorsWorked example
y = −1/x² and y = −4/x²
Both graphs lie below the x-axis.
At x = ±2: −1/4 and −1
The second curve lies lower on both sides.
−4/x² = 4(−1/x²)
A vertical stretch by factor 4 preserves the asymptotes.
04 / Sketch and recover
Use an asymptote and a point on each branch.
Mark the excluded input, draw the asymptotes, decide the sign on each side and calculate a few easy points. Draw smooth branches approaching the asymptotes. Do not join the branches across the undefined input.
To recover k from a known point (p, q), use k = pq for k/x, or k = p²q for k/x². Here p must be non-zero.
Which reciprocal fits the point?Worked example
The graph y = k/x² passes through (−2, 5)
Substitute both coordinates.
5 = k/4 ⇒ k = 20
Therefore y = 20/x².
At x = 2: y = 5
The matching point on the other branch follows from symmetry.
Range: y > 0
All positive values occur; zero does not.
05 / Move the asymptotes
The denominator tells you the forbidden input.
For y = k/(x − h) + v, with k ≠ 0, the vertical asymptote is x = h and the horizontal asymptote is y = v. The domain excludes h and the range excludes v.
y = 4/(x + 2) − 1
Asymptotes: x = −2, y = −1
Find intercepts from the equation: x = 0 gives y = 1, while y = 0 gives x = 2. The graph may cross the coordinate axes even though it never crosses its own asymptotes.
For a squared denominator, y = k/(x − h)² + v stays above v if k > 0 and below v if k < 0. The detailed transformation rules appear in the graph transformations lesson.
A shifted squared reciprocalWorked example
y = 8/(x − 1)² − 2
Asymptotes: x = 1 and y = −2.
At x = 0: y = 6
The y-intercept is (0, 6).
0 = 8/(x − 1)² − 2
(x − 1)² = 4
For x-intercepts, solve with the domain restriction x ≠ 1.
x = −1 or 3
Two crossings: (−1, 0), (3, 0). Range: y > −2.
06 / Your turn
Keep the denominator restriction in every answer.
State the branches, asymptotes and useful points clearly.
02 · Squared denominator
y = −12/x²
Hint
x² is positive on the domain.
Worked solution
Both branches lie below the axis, symmetric in the y-axis. Points (−2, −3), (2, −3). Asymptotes x = 0, y = 0; domain x ≠ 0; range y < 0.
03 · Find the numerator
The graph y = k/x passes through (−3, 4). Find k and the output at x = 6.
Hint
k = xy at a known point.
Worked solution
k = −3 · 4 = −12
At x = 6: y = −12/6 = −2
04 · Compare two branches
Compare 5/x and 1/x for x > 0 and x < 0.
Hint
Subtract the outputs.
Worked solution
5/x − 1/x = 4/x
The difference is positive for x > 0, so 5/x is above. It is negative for x < 0, so 5/x is below.
05 · Translated curve
y = 6/(x − 2) + 3
Hint
Find the asymptotes before the intercepts.
Worked solution
Asymptotes: x = 2, y = 3
At x = 0: y = 0
At y = 0: x = 0
Both intercepts are the origin. Domain x ≠ 2; range y ≠ 3. Relative to its centre (2, 3), the branches lie lower-left and upper-right.
06 · Two crossings
y = 18/(x + 1)² − 2
Hint
Set y = 0 and solve the squared equation.
Worked solution
(x + 1)² = 9
x = −4 or 2
The y-intercept is (0, 16). Asymptotes: x = −1 and y = −2. Domain x ≠ −1; range y > −2.
07 / Recap
A branch and an asymptote are different objects.
- Exclude zero denominators.
- A squared denominator keeps the same sign on both sides.
- For non-zero k, the basic asymptotes are x = 0 and y = 0.
- Use a known point to recover k.
- Translate the asymptotes along with the curve.
Section 1 of 7 · The two families