A rational function y=Q(x)P(x) has vertical asymptotes where Q(x)=0, and its behaviour for large ∣x∣ (horizontal or oblique asymptote) comes from dividing. Writing y=f(x) as a quadratic in x and using the discriminant gives the range and turning points.
Cambridge 9231 Paper 1AQA 7367 Paper 1AQA 7367 Paper 2≈ 40 minRevision 1 · updated 2026-10-27
Use the discriminant to find the range and turning points.
Pause and think
What happens to y=x−32x+1 when x is a million?
Section 2 of 14
Core idea
Start hereA curve that runs from a wall
Graph y=x−32x+1 and it shoots up near x=3 like a wall it can't touch, while far away it flattens out at height 2. Those invisible guides, asymptotes, are what you find first when sketching any rational function.
The big idea
Vertical asymptotes where the denominator is zero; long-run behaviour from division (horizontal or oblique asymptote); the range from the discriminant of y=f(x) treated as an equation in x.
1. Asymptotes
Horizontal asymptote of cx+dax+b
y=c+xda+xb
Divide top and bottom by x.
→ca as ∣x∣→∞
xb,xd→0.
Worked exampleOblique asymptote
Find the asymptotes of y=x+1x2+3x+4.
Goal
All asymptotes.
Strategy
Denominator zero; polynomial division.
1
x=−1
Vertical.
2
x2+3x+4=(x+1)(x+2)+2
Divide.
3
y=x+2+x+12
Remainder term →0.
Answerx=−1 and y=x+2.
✓ Checkx=100: y=10110304=102.02≈102 ✓.
2. Range and turning points
Discriminant method
y=x−3x2−5⇒yx−3y=x2−5
Multiply out.
x2−yx+(3y−5)=0
Quadratic in x.
y2−4(3y−5)≥0
Real x exists only if the discriminant is ≥0.
(y−2)(y−10)≥0⇒y≤2 or y≥10
Solve in y.
The boundary values 2 and 10 are the y-coordinates of the turning points.
Worked exampleGreatest value of x2+1x
Find the range.
Goal
Range.
Strategy
Discriminant.
1
yx2−x+y=0
Rearrange.
2
1−4y2≥0
Discriminant.
3
−21≤y≤21
Solve.
AnswerRange [−21,21]; greatest value 21 at x=1.
✓ Check1+11=21 ✓.
Sketching checklist
Drawing diagram…
Each feature usually earns a mark.
🔍 Edge caseCrossing a horizontal asymptoteAllowed: y=x2+1x crosses y=0 at the origin.
Summary
Vertical: denominator zero.
Divide for long-run behaviour.
Discriminant gives range.
Key words
Oblique asymptote
A slanted line approached for large |x|.
📜 HistoryAsymptote.The word comes from Greek "not falling together". Apollonius used it around 200 BC for the lines a hyperbola approaches but never meets.
🤯 Fun factCrossing an asymptote.Graphs can cross horizontal or oblique asymptotes (often in the middle). They just can't cross vertical ones, because the function isn't defined there.
Section 3 of 14
Key formulae
Linear over lineary=cx+dax+b:x=−cd,y=ca
Oblique asymptotex+rx2+px+q=x+m+x+rk⇒y=x+m
Range via discriminanty=f(x)→Ax2+Bx+C=0(in x),B2−4AC≥0
🪄 TrickTest just beside each asymptote.Plug in x=3.01 and x=2.99 (for an asymptote at x=3). The signs tell you which way the graph shoots off on each side.
Section 4 of 14
Worked examples
Example 1 · Oblique asymptote0 / 3 steps
Find the asymptotes of y=x+1x2+3x+4.
Example 2 · Range0 / 3 steps
Find the set of values taken by y=x−3x2−5.
Section 5 of 14
Explore the graph
📈Linear over linear
Loading the interactive graph…
Try this
Move d: the vertical asymptote is at x=−d.
Move a: the horizontal asymptote is y=a.
For which b does the curve collapse to a straight line? (When b=ad.)
y=x+dax+b has asymptotes x=−d and y=a.
🌍 In the real worldLens formula.v1=f1−u1 gives v=u−ffu: a rational function. Its vertical asymptote at u=f is why an object at the focal point forms no image.
Section 6 of 14
Common mistakes
IncorrectHorizontal asymptote y=db
Correcty=ca
For large x only the leading terms matter.
IncorrectGraph crossing a vertical asymptote
CorrectIt can't
The function is undefined there.
IncorrectSolving B2−4AC≥0 for x
CorrectIt's an inequality in y
You need the y-values for which some real x exists.
The curve: getting closer and closer
🫠
The asymptote: you will never actually reach me.
Section 7 of 14
Exam tip
🎯 Exam tipLabel everything.Sketch marks need asymptote equations written on the lines and intercept coordinates on the axes.