Rational function graphsAQA A-Level Further Maths: Revision notes

A rational function is a ratio of polynomials. For $y=\frac{ax+b}{cx+d}$ the graph is a hyperbola with two branches, drawn around two asymptotes: lines the curve gets arbitrarily close to but never reaches.

• Vertical asymptote: where the denominator is zero, $x=-\frac dc$ (provided the numerator is not also zero there).
• Horizontal asymptote: as $x\to\pm\infty$ the constants become negligible, so $y\to\frac ac$.
• Intercepts: $x$-axis where $ax+b=0$; $y$-axis at $x=0$, $y=\frac bd$. Example: $y=\frac{2x+1}{x-3}$ has asymptotes $x=3$ and $y=2$, and meets the axes at $\left(-\frac12,0\right)$ and $\left(0,-\frac13\right)$. Writing $y=2+\frac{7}{x-3}$ shows the curve lies above $y=2$ when $x>3$ and below it when $x<3$.

For $y=\frac{ax^2+bx+c}{dx^2+ex+f}$ (some coefficients may be zero), factorise where you can.

• Vertical asymptotes at each real root of the denominator.
• Horizontal asymptote $y=\frac ad$ when both degrees are 2; $y=0$ when the numerator has lower degree (for example $y=\frac{x}{x^2+1}$).
• Intercepts: numerator $=0$ for the $x$-axis, $x=0$ for the $y$-axis.
• Side of the asymptote: look at the sign of $y-\frac ad$ for large $|x|$. Example: $y=\frac{x^2}{x^2+2x-3}=\frac{x^2}{(x-1)(x+3)}$ has asymptotes $x=1$, $x=-3$ and $y=1$, and touches the $x$-axis at the origin because $x^2$ is a repeated factor. A curve can cross its horizontal asymptote: solve $y=\frac ad$ to find where (here $x=\frac32$). Oblique asymptotes are not needed at AS.

To find where a curve meets a straight line, equate the two expressions and multiply through by the denominator. The result is usually a quadratic whose roots are the $x$-coordinates of the intersections. Its discriminant tells you how many intersections there are: $b^2-4ac>0$ gives two, $=0$ gives one (a tangent), $<0$ gives none. Example: $\frac{x+3}{x-2}=5x-9\Rightarrow x+3=5x^2-19x+18\Rightarrow x^2-4x+3=0$, so $x=1$ or $3$, giving the points $(1,-4)$ and $(3,6)$.

To solve $\frac{p(x)}{q(x)}>g(x)$, never multiply by $q(x)$: it may be negative. Either multiply by $q(x)^2$, which is always positive, or move everything to one side and combine into a single fraction. Then find the critical values (zeros of the numerator and of the denominator) and test the sign in each interval. Example: $\frac{3x}{x+2}<x\Rightarrow\frac{x(x-1)}{x+2}>0$. Critical values are $-2$, $0$ and $1$; the expression is positive for $-2<x<0$ and for $x>1$. Asymptote values are never included in the solution because the function is undefined there.

To find the values a function can take, write $y=\frac{p(x)}{q(x)}$, rearrange into a quadratic in $x$ with $y$ in the coefficients, and require the discriminant to be non-negative (so that a real $x$ exists). The values of $y$ at the boundary give repeated roots, which are the stationary points. Example: $y=\frac{x+1}{x^2+3}\Rightarrow yx^2-x+3y-1=0$. Real $x$ needs $1-4y(3y-1)\ge0\Rightarrow12y^2-4y-1\le0\Rightarrow(6y+1)(2y-1)\le0$, so $-\frac16\le y\le\frac12$. At $y=\frac12$: $(x-1)^2=0$, giving $\left(1,\frac12\right)$. At $y=-\frac16$: $(x+3)^2=0$, giving $\left(-3,-\frac16\right)$.