Practice Maths

Reciprocal Functions

Key Terms

The basic reciprocal function is y = 1/x (a hyperbola with two branches).
Vertical asymptote
x = 0 (the y-axis) — the curve approaches but never touches it.
Horizontal asymptote
y = 0 (the x-axis) — the curve approaches it as x → ±∞.
The general form is y = a/(x − h) + k:
    — Vertical asymptote: x = h
    — Horizontal asymptote: y = k
    — a > 0: branches in quadrants 1 and 3 (relative to centre of symmetry)
    — a < 0: branches in quadrants 2 and 4
The centre of symmetry (intersection of asymptotes) is at (h, k).
Domain: all x ≠ h   Range: all y ≠ k
FunctionVert. asymptoteHoriz. asymptoteDomain / Range
y = 1/xx = 0y = 0x ≠ 0 / y ≠ 0
y = 1/(x − 2)x = 2y = 0x ≠ 2 / y ≠ 0
y = 1/x + 3x = 0y = 3x ≠ 0 / y ≠ 3
y = 2/(x − 1) + 4x = 1y = 4x ≠ 1 / y ≠ 4

Graph of y = 1/x and y = 2/(x − 1) + 2

x y 1 −1 1 −1 y=1/x y=2/(x−1)+2 x=1 y=2
Hot Tip To find intercepts of y = a/(x−h) + k: for the y-intercept, set x = 0 and calculate. For the x-intercept, set y = 0 → a/(x−h) = −k → x−h = −a/k → x = h − a/k (provided k ≠ 0).

Worked Example 1 — Key features of a hyperbola

Question: For y = 3/(x + 2) − 1, state the asymptotes, domain, range, and intercepts.

Asymptotes: Vertical: x = −2   Horizontal: y = −1

Domain: x ≠ −2   Range: y ≠ −1

y-intercept (x = 0): y = 3/2 − 1 = 1/2 → (0, 1/2)

x-intercept (y = 0): 0 = 3/(x+2) − 1 → 1 = 3/(x+2) → x+2 = 3 → x = 1 → (1, 0)

Worked Example 2 — Behaviour as x → ±∞

Question: Describe the behaviour of y = 2/(x − 3) + 1 as x → ∞ and x → −∞.

As x → +∞: 2/(x−3) → 0+, so y → 1+ (approaches y = 1 from above)

As x → −∞: 2/(x−3) → 0−, so y → 1− (approaches y = 1 from below)

Near x = 3 from right: 2/(x−3) → +∞, so y → +∞

Near x = 3 from left: 2/(x−3) → −∞, so y → −∞

Why the Shape of a Hyperbola?

Understanding y = 1/x means understanding how the quotient behaves. When x is very large (say x = 1000), y = 1/1000 is tiny — close to zero but positive. When x is very small and positive (say x = 0.001), y = 1/0.001 = 1000 is enormous. As x approaches zero from the positive side, y shoots upward toward +∞. As x approaches zero from the negative side, y shoots downward toward −∞.

What x can never equal is zero, because 1/0 is undefined. And y can never equal zero, because 1/x = 0 has no solution. This gives two asymptotes: the vertical asymptote x = 0 (the y-axis) and the horizontal asymptote y = 0 (the x-axis). The curve has two branches — one in the first quadrant (both x and y positive) and one in the third quadrant (both negative).

Asymptotes: Lines the Curve Approaches But Never Reaches

An asymptote is a line that the curve approaches as x or y goes to infinity (or as x approaches a specific value). For y = a/(x − h) + k:

The vertical asymptote is x = h, found by setting the denominator equal to zero: x − h = 0. The curve becomes unbounded near this line because you are dividing by a number approaching zero.

The horizontal asymptote is y = k, found by considering what happens as x → ±∞. The fraction a/(x−h) → 0, so y → k. The curve approaches but never reaches y = k because a/(x−h) = 0 has no solution.

The asymptotes must NEVER be crossed in a sketch. Draw them as dashed lines and ensure the curve curves toward them without touching.

Transformations of y = 1/x

Every feature of y = a/(x − h) + k can be understood as a transformation of the basic y = 1/x:

The parameter h shifts the graph horizontally: replacing x with (x − h) moves the vertical asymptote from x = 0 to x = h (shift right h units if h > 0).

The parameter k shifts the graph vertically: adding k moves the horizontal asymptote from y = 0 to y = k.

The parameter a stretches the graph vertically (if |a| > 1) or compresses it (if 0 < |a| < 1). If a < 0, it also reflects the graph in the x-axis, swapping which quadrants (relative to the asymptote intersection) contain each branch.

Domain and Range: Everything Except the Asymptotes

For y = a/(x − h) + k, the domain excludes only the vertical asymptote: domain is all real x except x = h. The range excludes only the horizontal asymptote: range is all real y except y = k. Both domain and range consist of two intervals separated by the excluded value.

To find intercepts: the y-intercept is found by setting x = 0 (provided 0 is in the domain, i.e., h ≠ 0); the x-intercept is found by setting y = 0 and solving. For y = 2/(x−3) + 1: set y = 0: 2/(x−3) = −1, so x − 3 = −2, x = 1. x-intercept at (1, 0).

Two Branches and Quadrant Positions

A hyperbola always has two separate branches. For the basic y = 1/x (a > 0), one branch is in the first quadrant (x > 0, y > 0) and one is in the third quadrant (x < 0, y < 0). For y = −1/x (a < 0), one branch is in the second quadrant (x < 0, y > 0) and one in the fourth (x > 0, y < 0). After transformation, “quadrant” is relative to the intersection of the asymptotes (h, k).

Exam Tip: The asymptotes are NEVER actually reached by the curve. Never draw the curve touching or crossing a dashed asymptote line. In exam sketches, the hyperbola branches should curve toward each asymptote, getting infinitely close but never touching. This is one of the most common sketch errors that costs presentation marks.
Exam Tip: The vertical asymptote is where the DENOMINATOR equals zero, not where the numerator equals zero. For y = 3/(2x − 6), the vertical asymptote is where 2x − 6 = 0, so x = 3. The x-intercept (where y = 0) requires the numerator to equal zero, but since the numerator is 3 (a constant), there is no x-intercept for this function.

Mastery Practice

See Answers ➔
  1. Fluency

    State the vertical asymptote, horizontal asymptote, domain and range of each function.

    1. (a) y = 1/(x − 3)
    2. (b) y = 1/x + 2
    3. (c) y = 4/(x + 1) − 3
    4. (d) y = −2/(x − 5) + 1
  2. Fluency

    Find the x-intercept and y-intercept (where they exist) for each function.

    1. (a) y = 1/x + 2
    2. (b) y = 3/(x − 2) + 1
    3. (c) y = −2/(x + 3)
    4. (d) y = 5/(x − 1) − 5
  3. Fluency

    Evaluate each function at the given x-value.

    1. (a) f(x) = 2/x at x = −4
    2. (b) g(x) = 1/(x − 1) + 3 at x = 3
    3. (c) h(x) = −4/(x + 2) + 1 at x = 2
  4. Fluency

    Describe the transformations from y = 1/x to each function.

    1. (a) y = 1/(x − 4)
    2. (b) y = 1/x − 3
    3. (c) y = −1/x + 2
    4. (d) y = 3/(x + 2) + 1
  5. Understanding

    Sketching hyperbolas.

    Sketch each function: draw dashed asymptotes, plot the intercepts, and draw both branches of the hyperbola.
    1. (a) y = 2/(x − 1) + 3 — state all key features before sketching
    2. (b) y = −1/(x + 2) − 1 — state all key features before sketching
  6. Understanding

    Writing the equation of a hyperbola from its features.

    Method: The asymptotes give h and k. Write y = a/(x−h)+k, then substitute a known point to find a.
    1. (a) Vertical asymptote x = 2, horizontal asymptote y = −1, passes through (3, 1).
    2. (b) Asymptotes x = −3 and y = 4, passes through (−2, 1).
    3. (c) Asymptotes x = 0 and y = 0, passes through (2, −3) (a multiple of y = 1/x).
  7. Understanding

    Solving equations involving reciprocal functions.

    1. (a) Solve 3/(x − 1) = 6.
    2. (b) Solve 2/(x + 3) + 1 = 5.
    3. (c) Solve 4/(x − 2) = 2/(x + 1). (Cross-multiply to solve.)
  8. Understanding

    Rewriting in the form y = a/(x − h) + k.

    Method: For a rational function like y = (2x + 5)/(x + 1), divide to separate into quotient + remainder form.
    1. (a) Rewrite y = (3x + 7)/(x + 2) in the form y = a/(x − h) + k. State asymptotes.
    2. (b) Rewrite y = (2x − 3)/(x − 1) in the form y = a/(x − h) + k. State asymptotes.
  9. Problem Solving

    Modelling with reciprocal functions.

    Challenge. The time T (hours) to complete a job with n workers is modelled by T = 12/n, where n ≥ 1.
    1. (a) How long does the job take with 4 workers? With 8 workers?
    2. (b) Sketch T against n for n = 1 to 12. Label two points and describe the shape.
    3. (c) How many workers are needed to complete the job in 1.5 hours? Is the answer a whole number? What practical value should be recommended?
  10. Problem Solving

    Intersecting a hyperbola with a line.

    Challenge.
    1. (a) Find the coordinates where y = 2/(x − 1) + 3 intersects the line y = 5.
    2. (b) Find where y = 6/x intersects y = x + 1. (Form a quadratic and solve.)
    3. (c) Determine whether the line y = 2x + 3 ever intersects y = 4/x. Find the intersection points or prove there are none.