Composite transformations of curvesAQA A-Level Further Maths: Revision notes

For a curve $y=\mathrm{f}(x)$:

• $y=\mathrm{f}(x)+a$ is a translation $\begin{pmatrix} 0 \\ a \end{pmatrix}$; $y=\mathrm{f}(x+a)$ is a translation $\begin{pmatrix} -a \\ 0 \end{pmatrix}$.
• $y=a\mathrm{f}(x)$ is a stretch with scale factor $a$ parallel to the $y$-axis; $y=\mathrm{f}(ax)$ is a stretch with scale factor $\frac{1}{a}$ parallel to the $x$-axis.
• $y=-\mathrm{f}(x)$ is a reflection in the $x$-axis; $y=\mathrm{f}(-x)$ is a reflection in the $y$-axis.

Changes to $x$ happen inside the brackets and act 'backwards'; changes to $y$ act as you expect. Asymptotes, intercepts and turning points move with the curve.

A rotation about the origin maps a point as follows:

• $90^\circ$ anticlockwise: $(x,y)\to(-y,x)$
• $90^\circ$ clockwise: $(x,y)\to(y,-x)$
• $180^\circ$: $(x,y)\to(-x,-y)$

To find the image of $y=\mathrm{f}(x)$, let $(X,Y)$ be the image of $(x,y)$, write $x$ and $y$ in terms of $X$ and $Y$, and substitute into $y=\mathrm{f}(x)$. For $90^\circ$ anticlockwise, $X=-y$ and $Y=x$, so $-X=\mathrm{f}(Y)$. Writing the image in terms of $x$ and $y$ again:

• $90^\circ$ anticlockwise: $x=-\mathrm{f}(y)$
• $90^\circ$ clockwise: $x=\mathrm{f}(-y)$
• $180^\circ$: $y=-\mathrm{f}(-x)$

Example: $y=x^2+1$ rotated $90^\circ$ anticlockwise gives $x=-(y^2+1)$. Check: $(1,2)\to(-2,1)$ and $-2=-(1+1)$.

An enlargement with scale factor $k$ and centre the origin maps $(x,y)\to(kx,ky)$. The image of $y=\mathrm{f}(x)$ satisfies $\frac{y}{k}=\mathrm{f}\left(\frac{x}{k}\right)$, so $y=k\,\mathrm{f}\left(\frac{x}{k}\right).$ It is a stretch of factor $k$ in the $y$-direction and a stretch of factor $k$ in the $x$-direction. Example: $y=x^2+1$ with $k=2$ gives $y=2\left(\frac{x^2}{4}+1\right)=\frac{x^2}{2}+2$. Check: $(1,2)\to(2,4)$ and $\frac{4}{2}+2=4$.

An enlargement with scale factor $-1$ is the same as a rotation through $180^\circ$ about the origin.

A composite transformation is two or more transformations applied one after another. '$A$ followed by $B$' means do $A$ first.

Work one step at a time: write the equation after the first transformation, then apply the second to that equation, not to the original.

Example: stretch $y=\mathrm{e}^x$ by factor 3 parallel to the $y$-axis, then translate by $\begin{pmatrix} 0 \\ -1 \end{pmatrix}$: $y=3\mathrm{e}^x$ and then $y=3\mathrm{e}^x-1$. In the opposite order, $y=\mathrm{e}^x-1$ and then $y=3(\mathrm{e}^x-1)=3\mathrm{e}^x-3$, which is a different curve.

Order matters whenever one step is a translation and the other is a stretch, an enlargement or a rotation, because the second step also acts on the shift made by the first. Two translations, or two stretches in the same direction, can be done in either order.

To find where a feature ends up, apply the transformations to it in order. Asymptotes are lines, so transform them like points on the line.

Example: $y=\frac{1}{x-2}$ has asymptotes $x=2$ and $y=0$. Enlarge with scale factor 2, then translate by $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$. The asymptote $x=2$ goes to $x=4$ and then $x=1$; the asymptote $y=0$ stays $y=0$ and then becomes $y=1$. The image is $y=\frac{4}{x-1}+1$.

Worked example: rotate $y=\ln x$ through $90^\circ$ anticlockwise: $x=-\ln y$, so $y=\mathrm{e}^{-x}$. Check: $(\mathrm{e},1)\to(-1,\mathrm{e})$ and $\mathrm{e}^{1}=\mathrm{e}$. A translation $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ then gives $y=\mathrm{e}^{-x}+2$, with asymptote $y=2$.