Reciprocal Graphs (AA SL)
Reciprocal graphs behave differently from anything else on the AA SL course: instead of a single smooth curve, they split into branches that hug invisible lines called asymptotes. This page focuses on finding those asymptotes and the other key features of \(\dfrac{1}{x}\)-style and rational graphs, building on the general transformation rules from Graphs & Transformations.
9 questions on this sub-topic.
Asymptotes of a shifted reciprocal
Covered under IB syllabus reference SL2.8, which covers the reciprocal function \(y=\dfrac{1}{x}\) and rational functions of the form \(y=\dfrac{ax+b}{cx+d}\), including their asymptotes. Neither result below is a formula-booklet entry - both follow from the general transformation rules.
Shifted reciprocal
\(y=\dfrac{a}{x-h}+k\)
Vertical asymptote \(x=h\) (denominator zero); horizontal asymptote \(y=k\) (the value the graph tends to as \(x\to\pm\infty\)).
Need the general translation and stretch rules first? See Graphs & Transformations.
Worked examples
Let \(f(x)=\dfrac{3}{x+2}-1\).
(a) State the vertical asymptote.
(b) State the horizontal asymptote.
(c) Find the \(y\)-intercept.
Worked solution
(a) Denominator zero: \(x+2=0\Rightarrow x=-2\). A1
(b) As \(x\to\pm\infty,\ f\to0-1=-1\), so \(y=-1\). A1
(c) \(f(0)=\dfrac{3}{2}-1=0.5\). A1
Let \(f(x)=\dfrac{x^2-x-6}{x-1}\).
(a) Find the vertical asymptote.
(b)(i) Find the \(x\)-intercept with \(x<0\).
(b)(ii) Find the \(x\)-intercept with \(x>0\).
(c) Find the \(y\)-intercept.
(d) State whether \(f\) has a horizontal or oblique asymptote.
Worked solution
(a) \(x=1.\) A1
(b)(i) \(x^2-x-6=0\Rightarrow(x-3)(x+2)=0,\) so \(x\) M1
\(=-2\) (negative root). A1
(b)(ii) \(x=3\) (positive root). A1
(c) \(f(0)=\dfrac{-6}{-1}=6,\) so \((0,6).\) A1
(d) Degree of numerator is one more than denominator, so there is an oblique asymptote. M1 A1
Common mistakes
- Getting the sign of the vertical asymptote backwards. For \(y=\dfrac{a}{x-h}+k\), the asymptote is \(x=h\), not \(x=-h\) - find it by setting the denominator to zero and solving, not by reading off the sign in the bracket.
- Forgetting the horizontal asymptote shifts too. Adding a constant \(k\) outside the fraction moves the horizontal asymptote from \(y=0\) to \(y=k\) - it's easy to state \(y=0\) out of habit even after a vertical shift.
- Assuming every rational graph has a horizontal asymptote. That's only true when the numerator's degree is less than or equal to the denominator's. If the numerator's degree is exactly one higher, the asymptote is oblique (slanted), not horizontal.
Ready to practise properly?
9 reciprocal-graph questions, marked instantly like the real exam.
Quick answers
How do I find the asymptotes of a shifted reciprocal graph?
For \(y=\dfrac{a}{x-h}+k\), the vertical asymptote is \(x=h\) (where the denominator is zero) and the horizontal asymptote is \(y=k\) (the value the graph approaches as \(x\to\pm\infty\)).
When does a rational function have an oblique asymptote instead of a horizontal one?
When the degree of the numerator is exactly one greater than the degree of the denominator, the graph approaches a slanted (oblique) line rather than a horizontal one, found by polynomial division.