Sketching graphs of functionsEdexcel International A Level Maths: Revision notes

A cubic $y=ax^3+bx^2+cx+d$ with $a>0$ rises from bottom left to top right ($y\to-\infty$ as $x\to-\infty$, $y\to+\infty$ as $x\to+\infty$); with $a<0$ it falls. Sketch it by finding:

• the $x$-intercepts, by factorising: $y=x(x-2)(x+2)$ crosses at $-2$, $0$ and $2$;
• the $y$-intercept, by putting $x=0$;
• whether each root is a crossing or a touching point. A repeated root such as $x^2$ in $y=x^2(x-4)$ makes the curve touch the axis at that point (a turning point) rather than cross it; a single root crosses.

The graph of $y=\frac kx$ ($x\neq0$) has two branches: in the first and third quadrants if $k>0$, and in the second and fourth if $k<0$. It is symmetrical about $y=x$ (for $k>0$) and never touches the axes. The graph of $y=\frac k{x^2}$ ($x\neq0$) lies above the $x$-axis for $k>0$, in the first and second quadrants, and is symmetrical about the $y$-axis. For $k<0$ it lies below the axis. An asymptote is a line that the curve approaches but never reaches. Both $\frac kx$ and $\frac k{x^2}$ have asymptotes $x=0$ and $y=0$.

For $0^\circ\le x\le360^\circ$:

• $y=\sin x$: starts at $(0^\circ,0)$, maximum $(90^\circ,1)$, crosses at $180^\circ$, minimum $(270^\circ,-1)$, ends at $0$. Period $360^\circ$.
• $y=\cos x$: starts at $(0^\circ,1)$, crosses at $90^\circ$, minimum $(180^\circ,-1)$, crosses at $270^\circ$, ends at $1$.
• $y=\tan x$: passes through the origin and $(180^\circ,0)$, increasing, with vertical asymptotes at $x=90^\circ$ and $x=270^\circ$. Period $180^\circ$; range all real numbers.

The $x$-coordinates where the graphs of $y=f(x)$ and $y=g(x)$ meet are the solutions of $f(x)=g(x)$. A sketch shows how many solutions there are before you solve algebraically. Example: $\sin x=\cos x$ in $0^\circ\le x\le360^\circ$: the curves cross twice, at $45^\circ$ and $225^\circ$ (from $\tan x=1$). Example: $x(x-2)(x+2)=\frac2x$. The cubic is negative for $0<x<2$ while $\frac2x$ is positive, so any intersection for $x>0$ has $x>2$. By symmetry there are two solutions in total, $x=\pm\sqrt{2+\sqrt6}$.

A good sketch shows the overall shape, all intercepts with coordinates, turning points where they touch an axis, and any asymptotes (drawn as dashed lines and labelled with equations). It does not need to be to scale. For a curve such as $y=x^2(x-4)$: touches the $x$-axis at the origin, crosses at $(4,0)$, passes below the axis for $x<4$ and rises to $+\infty$ for $x>4$. So the line $y=5$ meets it once.