1.2 Rational functions and graphs
- Syllabus
- 9231–2028–2029
- Topic
- 1.2
- Level
- AS
Learning objectives
1.2.1Rational graphs• sketch graphs of simple rational functions, including the determination of oblique asymptotes, in cases where the degree of the numerator and the denominator are at most 2 Including determination of the set of values taken by the function, e.g. by the use of a discriminant. Detailed plotting of curves will not be required, but sketches will generally be expected to show significant features, such as turning points, asymptotes and intersections with the axes.1.2.2Related graphs• understand and use relationships between the graphs of y = f(x), y2 = f(x), y x 1 f= ^h, y xf= ^h and y xf= ^h. Including use of such sketch graphs in the course of solving equations or inequalities.
Let the algebra determine every feature of a rational sketch
For y=P(x)/Q(x) with numerator and denominator degrees at most 2: factor first; record excluded x-values and cancelled-factor holes; find axis intercepts; divide P by Q; locate vertical and end-behaviour asymptotes; use y′ or the discriminant range calculation for turning values; then use signs and limits on each interval to join the features with the correct branches.
After cancellation, $Q(a)=0$ gives a vertical asymptote $x=a$ when the numerator is non-zero there. If $\deg P=\deg Q$, the horizontal asymptote is the ratio of leading coefficients. If $\deg P=\deg Q+1$, division gives\frac{P(x)}{Q(x)}=m x+c+\frac{R(x)}{Q(x)},so $y=m x+c$ is the oblique asymptote.
To find the range, treat y as a fixed candidate and rearrange yQ(x)=P(x) into a quadratic in x. A y-value occurs exactly when this equation has an allowed real x, so require discriminant Δ≥0 and then remove values produced only by an excluded x. Equality Δ=0 usually identifies a turning value.
Fory=\frac{x^2-2x+3}{x^2+1},thedenominatorisalwayspositiveand(y-1)x^2+2x+(y-3)=0.Hence\Delta=4-4(y-1)(y-3)\ge0\iff y^2-4y+2\le0,sotherangeis2-\sqrt2\le y\le2+\sqrt2.The horizontal asymptote $y=1$ is crossed at $x=1$; an asymptote is limiting behaviour, not automatically a forbidden y-value.
A cancelled denominator factor makes a hole, not a vertical asymptote. A feature-complete sketch must show significant intercepts, turning points and asymptotes; detailed point plotting is neither required nor a substitute for those deductions.
Transform a known graph by mapping its branches and points
| Required graph | Construction from y=f(x) | Features to track |
|---|---|---|
| y2=f(x) | Keep only where f(x)≥0; replace each (x,f(x)) by (x,±f(x)) | Symmetric about x-axis; branches meet at zeros of f |
| y=1/f(x) | Replace each defined non-zero output y by 1/y | Same sign as f; zeros of f become vertical asymptotes; f→±∞ gives reciprocal output tending to 0 |
| y=∣f(x)∣ | Keep positive parts; reflect every negative part in the x-axis | Same domain and zeros; a simple crossing usually becomes a sharp minimum |
| y=f(∣x∣) | Keep the original half for x≥0 and reflect it in the y-axis; discard the original x<0 half | Always even; the right-hand domain/features determine both sides |
∣f(x)∣ changes the output, so it folds below-axis pieces upward. f(∣x∣) changes the input, so it copies the right-hand half to the left. For y2=f(x), negative f-values produce no real points; for 1/f(x), points where the original graph is undefined remain outside the domain even if the reciprocal limit is 0.
Use the transformed sketch to solve: an equation asks for x-coordinates of intersections with the relevant horizontal or comparison curve; an inequality asks for x-intervals where one graph lies above/below another. Mark critical x-values first, test strict versus inclusive endpoints, and exclude every undefined input.
These are not the generic translations f(x−a) or f(x)+a. Preserve the exact operation shown: squaring y, reciprocating the output, taking the output modulus and taking the input modulus produce four different domains and symmetries.