Graphs of functions, asymptotes and proportional relationshipsEdexcel A-Level Maths: Revision notes

To sketch a cubic or quartic: (1) find where it meets the $y$-axis ($x=0$); (2) find where it meets the $x$-axis by solving $y=0$; (3) decide the end behaviour from the sign and degree of the leading term. At a single root the curve crosses the axis. At a repeated root, such as $x^2$ in the factor, the curve touches the axis and turns round. End behaviour: a positive cubic goes from bottom left to top right ($y\to-\infty$ as $x\to-\infty$, $y\to\infty$ as $x\to\infty$); a positive quartic rises at both ends. A negative leading coefficient reflects these in the $x$-axis. Example: $y=x^2(2x-1)^2$ touches the $x$-axis at $x=0$ and $x=\frac12$, is never negative, and rises at both ends.

The graph of $y=\frac ax$ has two branches. For $a>0$ they lie in the first and third quadrants; for $a<0$ in the second and fourth. The axes are asymptotes: lines the curve approaches but never meets, $x=0$ and $y=0$. The graph of $y=\frac{a}{x^2}$ is always on one side of the $x$-axis, because $x^2>0$: above it if $a>0$, below if $a<0$. It is symmetric about the $y$-axis and has the same asymptotes. Both axes are asymptotes because $x$ can never be $0$ (so $y$ is undefined there) and $y$ can never be $0$ (a fraction with non-zero numerator is never $0$).

The graph of $y=\frac{a}{x+p}+q$ is $y=\frac ax$ translated $p$ units left and $q$ units up. Its asymptotes are $x=-p$ and $y=q$. Example: $y=\frac{3}{x-2}+1$ has asymptotes $x=2$ and $y=1$. It meets the $y$-axis where $x=0$: $y=-\frac32+1=-\frac12$. It meets the $x$-axis where $\frac{3}{x-2}=-1$, so $x=-1$. The curve cannot meet its own horizontal asymptote: setting $\frac{a}{x+p}+q=q$ gives $\frac{a}{x+p}=0$, which has no solution.

The solutions of $f(x)=g(x)$ are the $x$-coordinates of the points where the graphs $y=f(x)$ and $y=g(x)$ meet. The number of intersections is the number of real solutions. To find them exactly, set the two expressions equal and solve. Example: the curve $y=x^2(x-4)$ and the line $y=x-4$ meet where $x^2(x-4)=x-4$, so $(x-4)(x^2-1)=0$ and $x=4,\ 1,\ -1$. The points are $(4,0),\ (1,-3),\ (-1,-5)$. To solve $\frac{3}{x-2}+1=x+1$, simplify to $3=x(x-2)$, so $x=3$ or $x=-1$, giving the points $(3,4)$ and $(-1,0)$.

Direct proportion: $y\propto x$ means $y=kx$ for a constant $k$. The graph of $y$ against $x$ is a straight line through the origin with gradient $k$. Example: $C=kd$. Inverse proportion: $y\propto\frac1x$ means $y=\frac kx$, a reciprocal curve with asymptotes on both axes. Plotting $y$ against $\frac1x$ gives a straight line through the origin with gradient $k$. Example: $P$ is inversely proportional to $V$ with $P=150$ when $V=2$. Then $k=300$ and $P=\frac{300}{V}$; at $V=5$, $P=60$.