Sketching polynomial and reciprocal graphsAQA A-Level Maths: Revision notes

The end behaviour of a polynomial is decided by its highest-power term. For $y=ax^n$ as the leading term:

• odd $n$, $a>0$: $y\to\infty$ as $x\to\infty$ and $y\to-\infty$ as $x\to-\infty$ (cubic rising left to right);
• odd $n$, $a<0$: the reverse (falling left to right);
• even $n$, $a>0$: $y\to\infty$ at both ends (a $\cup$ or W shape);
• even $n$, $a<0$: $y\to-\infty$ at both ends (a $\cap$ shape). For example $y=3x^2-x^3$ has leading term $-x^3$, so it falls from the top left to the bottom right.

For a polynomial in factorised form, set $y=0$ for the roots (the $x$-intercepts) and set $x=0$ for the $y$-intercept.

• A single factor $(x-a)$: the curve crosses the $x$-axis at $x=a$.
• A repeated factor $(x-a)^2$: the curve touches the axis at $x=a$ and turns round (a turning point on the axis).
• A cubed factor $(x-a)^3$: the curve flattens as it crosses, an inflection on the axis. For $y=(x+2)(x-1)^2$: crosses at $(-2,0)$, touches at $(1,0)$, and $y=(2)(-1)^2=2$ at $x=0$, so the $y$-intercept is $(0,2)$.

A reliable method for a polynomial sketch:

1. Factorise and find the roots, marking crossing or touching.
2. Find the $y$-intercept.
3. Use the leading term for the ends.
4. Join the points with a smooth curve, using the sign between roots to decide whether the curve is above or below the axis. For $y=(x-1)(x+3)(x-4)$, the roots are $-3,1,4$ and the $y$-intercept is $12$. The curve rises from the bottom left, crosses at $-3$, peaks, crosses at $1$, dips below the axis, crosses at $4$ and rises on. The sign is positive for $-3<x<1$ and for $x>4$, negative for $x<-3$ and for $1<x<4$. A quartic $y=x^2(x-2)(x+1)$ has a $\cup$-type shape (positive $x^4$): it touches at the origin and crosses at $-1$ and $2$.

For $a>0$:

• $y=\frac{a}{x}$ has two branches, in the first and third quadrants. It has rotational symmetry about the origin.
• $y=\frac{a}{x^2}$ has two branches in the first and second quadrants, because $x^2>0$. It is symmetric in the $y$-axis. Both have the asymptotes $x=0$ (vertical) and $y=0$ (horizontal): the curve approaches these lines but never meets them. $\frac{a}{x}=0$ has no solution, and $x=0$ makes $y$ undefined. If $a<0$ the graph reflects in the $x$-axis: $y=-\frac{3}{x}$ lies in the second and fourth quadrants, and $y=-\frac{3}{x^2}$ in the third and fourth.

Direct and inverse proportion give the standard shapes. Write the relationship with a constant $k$, then use a known pair of values to find $k$.

• $y\propto x$: $y=kx$, a straight line through the origin with gradient $k$.
• $y\propto x^2$: $y=kx^2$, a parabola with its vertex at the origin.
• $y\propto\frac1x$: $y=\frac{k}{x}$, a reciprocal graph.
• $y\propto\frac{1}{x^2}$: $y=\frac{k}{x^2}$. Example: $y$ is inversely proportional to $x$ and $y=3$ when $x=4$. Then $k=xy=12$, so $y=\frac{12}{x}$ and at $x=6$, $y=2$. A graph of $y$ against $x$ is a straight line through the origin only for direct proportion.

A sketch does not need an accurate scale, but it must show every key feature, labelled with coordinates or equations:

• all axis intercepts, and whether the curve crosses or touches the $x$-axis;
• turning points if they can be found (for example by differentiation);
• the behaviour as $x\to\pm\infty$;
• asymptotes, drawn as dashed lines and labelled with their equations. Two curves on one set of axes should be sketched in the same way, with any intersections justified by sign or end-behaviour arguments.