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IB Maths AA SL · Unit 2: Functions

IB Maths AA SL Rational and Reciprocal Functions Questions

Exam-style IB Maths AA SL rational and reciprocal functions questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.

Practise Rational and Reciprocal Functions questions → AA SL formula booklet

What you need to know

Vertical asymptotes from the denominator = 0; horizontal from the ratio of leading terms as x → ∞. Sketch the shape, don't rely on a GDC — Paper 1 is calc-free. Rational functions and asymptotes overview →

What's examined in AA SL rational and reciprocal functions

The question bank covers these rational and reciprocal functions question types (number of questions in brackets):

Rational and Reciprocal Functions worked examples

Worked example 1: Finding asymptotes and intercepts · easy

Let $f(x) = \frac{4x - 5}{2x + 3}$. Find the exact equations of both asymptotes and the exact coordinates of the intercepts.

Solution

1. Vertical asymptote: $2x + 3 = 0 \implies \mathbf{x = -1.5}$.

2. Horizontal asymptote: $y = \frac{4}{2} = \mathbf{y = 2}$.

3. $y$-intercept: $f(0) = -\frac{5}{3} \implies \mathbf{\left(0, -\frac{5}{3}\right)}$.

4. $x$-intercept: $4x - 5 = 0 \implies \mathbf{(1.25, 0)}$.

Examiner tip: You can verify with a GDC graph, but exact values are best (and fastest) found algebraically.

Worked example 2: Inverse of a rational function · medium

Find the inverse function $f^{-1}(x)$ for $f(x) = \frac{3x + 4}{2x - 1}$, where $x \neq 0.5$.

Solution

1. Swap: $x = \frac{3y+4}{2y-1}$.

2. Multiply: $x(2y - 1) = 3y + 4$.

3. Expand: $2xy - x = 3y + 4$.

4. Group $y$: $2xy - 3y = x + 4 \implies y(2x - 3) = x + 4$.

5. Isolate: $\mathbf{f^{-1}(x) = \frac{x + 4}{2x - 3}}$.

Examiner tip: If you end up with all-negative signs (e.g. $\frac{-x-4}{3-2x}$), multiply top and bottom by $-1$ to match the examiner's mark scheme presentation.

Worked example 3: Partial fractions and translating reciprocals · hard

Let $f(x) = \frac{3x + 5}{x + 2}$ for $x \neq -2$. Show that $f(x)$ can be written as $A + \frac{B}{x+2}$, and describe the transformations mapping $y = \frac{1}{x}$ onto $y = f(x)$.

Solution

1. Split the numerator: $3x + 5 = 3(x + 2) - 1$.

2. Rewrite: $\frac{3(x+2) - 1}{x+2} = 3 - \frac{1}{x+2}$. So $\mathbf{A = 3}$, $\mathbf{B = -1}$.

3. Interpret: the $(x+2)$ shifts left by $2$, the $+3$ shifts up by $3$.

4. State Transformation 1: translation by vector $\binom{-2}{3}$.

5. Interpret the negative multiplier ($B = -1$) — this is a reflection.

6. State Transformation 2: reflection in the $x$-axis.

Examiner tip: Rewriting the numerator as a multiple of the denominator (e.g. $3x+5 = 3(x+2) - 1$) is safer than formal polynomial long division, which is prone to sign errors.

Try these IB Maths AA SL rational and reciprocal functions questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.

Question 1 · easy · 5 marks · Paper 1

The function \(f\) is defined by \(f(x) = \frac{1}{x-2} + 3\), for \(x \ne 2\).

  1. Write down the equations of the vertical and horizontal asymptotes of the graph of \(f\).

  2. Find the exact coordinates of the \(y\)-intercept.

  3. Find the exact coordinates of the \(x\)-intercept algebraically.

Attempt it and see the mark scheme →

Question 2 · medium · 5 marks · Paper 1

Consider the function \(f(x) = \frac{5x - 2}{x + 3}\), where \(x \ne -3\).

  1. Find an expression for the inverse function, \(f^{-1}(x)\).

  2. State the equations of the vertical and horizontal asymptotes of \(f^{-1}(x)\).

Attempt it and see the mark scheme →

Question 3 · hard · 6 marks · Paper 1

The curve \(C\) has equation \(y = \frac{2x+1}{x-1}\) for \(x \ne 1\). The straight line \(L\) has equation \(y = -x + k\), where \(k\) is a constant.

  1. Show that the \(x\)-coordinates of any points of intersection of \(C\) and \(L\) satisfy the equation \(x^2 + (1-k)x + (k+1) = 0\).

  2. Given that the line \(L\) is a tangent to the curve \(C\), use the discriminant to find the exact possible values of \(k\).

Attempt it and see the mark scheme →

All 32 rational and reciprocal functions questions with mark schemes →

FAQ

How many IB Maths AA SL rational and reciprocal functions questions are there?

There are 32 exam-style rational and reciprocal functions questions in the AA SL question bank (Paper 1: 25 · Paper 2: 7), graded 6 easy, 14 medium, 6 hard, 3 very hard, 3 starter. Every question has a full IB-style mark scheme (M, A and R marks).

Is rational and reciprocal functions on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 25 · Paper 2: 7. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.

Where can I get the mark schemes?

Open the AA SL Unit 2 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

More AA SL Unit 2 topics

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