Rational functions and their graphs
Sketch simple rational functions from their algebraic structure, identify asymptotes and use discriminants to find ranges.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Sketch rational functions with numerator and denominator degree at most 2
- Find vertical and oblique asymptotes and significant intercepts
- Use a discriminant to determine values taken by a function
- Relate transformed function graphs when solving equations or inequalities
Divide first, then inspect the remaining fraction
Polynomial division can reveal an oblique asymptote. For (x²+1)/(x−1), division gives x+1+2/(x−1), so x=1 is a vertical asymptote and y=x+1 is an oblique asymptote.
A sketch should show asymptotes, intercepts and turning points where they are straightforward to find. A cancelled factor is a removable hole, not a vertical asymptote; check the original domain.
A range question can become a quadratic condition
Set y=f(x), rearrange into a polynomial equation in x, and ask when it has at least one real solution. For a quadratic in x this means its discriminant is non-negative, with any excluded x-values checked separately.
For the example above, y=(x²+1)/(x−1) gives x²−yx+(y+1)=0. Its discriminant is y²−4y−4, so the range is y≤2−2√2 or y≥2+2√2.
Graph relationships can be read from the input or output: y=f(−x) reflects y=f(x) across the y-axis; y=f(|x|) mirrors the right half into the left; y²=f(x) gives y=±√f(x) where f(x)≥0; and y=f(1/x) uses reciprocal inputs and excludes x=0. For y=1/f(x), zeros of f become excluded inputs and may become vertical asymptotes.
Pause and explain
Which habit best checks your work on rational functions and their graphs?
Worked example
Find the asymptotes and range of f(x)=(x²+1)/(x−1).
Show the worked solution
- Division gives f(x)=x+1+2/(x−1), so x=1 and y=x+1 are asymptotes.
- Set y=f(x): x²−yx+y+1=0. It has a real x exactly when y²−4y−4≥0.
- Solve (y−2)²≥8, giving y≤2−2√2 or y≥2+2√2.
Answer Asymptotes x=1 and y=x+1; range (−∞, 2−2√2] ∪ [2+2√2, ∞).
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For f(x)=(x²−4)/(x−2), state the domain and describe the graph accurately.
Give me a hint
Factor the numerator, but retain the original restriction x≠2.
Compare my reasoning
- For x≠2, f(x)=x+2.
- The original function is undefined at x=2, so the point (2,4) is missing.
- The graph is the line y=x+2 with a removable hole, not a vertical asymptote.
Domain x≠2; line y=x+2 with a hole at (2,4).
Look for these in your work
- Original domain retained
- Cancellation explained
- Hole distinguished from asymptote
Find the range of g(x)=1/(x−2) for real x≠2.
Give me a hint
A reciprocal cannot equal zero; solve y=1/(x−2) for x when y≠0.
Compare my reasoning
- If y=1/(x−2), then x=2+1/y.
- This real x exists for every real y except y=0.
- Therefore the range excludes zero.
All real y except 0.
Look for these in your work
- Equation rearranged
- Zero excluded
- All non-zero values shown attainable
The graph y=f(x) has zeros at x=−3 and x=2. Where can y=1/f(x) have vertical asymptotes, and what domain issue must you mention?
Give me a hint
Zeros of f become denominator zeros in its reciprocal.
Compare my reasoning
- At x=−3 and x=2, f(x)=0, so 1/f(x) is undefined.
- These are vertical asymptotes if the zeros are isolated and the reciprocal grows without bound nearby.
- The reciprocal's domain excludes every zero of f; any other original exclusions must also be checked.
Potential vertical asymptotes at x=−3 and x=2; the reciprocal is undefined at zeros of f, and the local behaviour must confirm an asymptote.
Look for these in your work
- Zero-to-denominator connection
- Asymptote condition qualified
- Domain exclusions stated
Common mistakes
- Calling the quotient y=x+1 an exact curve instead of an asymptote
- Using a discriminant but forgetting to check the original domain
- Assuming every horizontal or oblique asymptote is reached by the graph
What can you explain now?
What does a non-cancelled zero of the denominator usually indicate?
Compare with the explanation
A vertical asymptote
The function is undefined there; check for a common factor before deciding whether there is an asymptote or a hole.
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Make it stick
How can polynomial division and a discriminant reveal a rational function's graph and range?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.