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

A translation slides a curve without turning or resizing it.

• $y=\mathrm{f}(x)+a$ translates by $\begin{pmatrix}0\\ a\end{pmatrix}$: up if $a>0$.
• $y=\mathrm{f}(x-a)$ translates by $\begin{pmatrix}a\\0\end{pmatrix}$: right if $a>0$. The sign inside the bracket works against intuition. Each point $(x,y)$ moves with the curve, and so do intercepts and asymptotes. For $y=x^2+2x$, translating by $\begin{pmatrix}3\\0\end{pmatrix}$ replaces $x$ by $x-3$: $y=(x-3)^2+2(x-3)=x^2-4x+3$.

A stretch changes distances from an axis by a scale factor.

• $y=a\,\mathrm{f}(x)$: stretch parallel to the $y$-axis, scale factor $a$. Point $(x,y)\to(x,ay)$.
• $y=\mathrm{f}(ax)$: stretch parallel to the $x$-axis, scale factor $\frac1a$. Point $(x,y)\to\left(\frac xa,y\right)$. So a stretch parallel to the $x$-axis with scale factor $k$ replaces $x$ by $\frac xk$; one parallel to the $y$-axis with scale factor $k$ replaces $y$ by $\frac yk$. Example: stretching $y=x^2+2x$ parallel to the $x$-axis with scale factor $2$ gives $y=\frac{x^2}{4}+x$. Intercepts on the axis you stretch away from move; intercepts on the other axis stay where they are.

• In the $x$-axis: $y=-\mathrm{f}(x)$; replace $y$ by $-y$.
• In the $y$-axis: $y=\mathrm{f}(-x)$; replace $x$ by $-x$.
• In the line $y=x$: swap $x$ and $y$, giving $x=\mathrm{f}(y)$.
• In the line $y=-x$: replace $x$ by $-y$ and $y$ by $-x$, giving $-x=\mathrm{f}(-y)$. Example: reflecting $y=x^2$ in $y=-x$ gives $-x=y^2$, a parabola opening to the left. Reflecting the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ in $y=x$ gives $\frac{y^2}{4}+\frac{x^2}{9}=1$.

For a curve given by an equation (a conic, a cubic, an ellipse), apply the substitution to every occurrence of $x$ or $y$, then simplify. The same rule moves asymptotes and intercepts: a translation by $\begin{pmatrix}a\\ b\end{pmatrix}$ moves an asymptote $x=k$ to $x=k+a$ and $y=m$ to $y=m+b$. Example: stretching the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ parallel to the $y$-axis with scale factor $\frac13$ replaces $y$ by $3y$, giving $\frac{x^2}{4}+y^2=1$, which meets the $y$-axis at $(0,\pm1)$.

A full description needs the type plus the details: translation with a vector, stretch with a direction (parallel to which axis) and scale factor, reflection with the mirror line. Compare the new equation with the old: with $\mathrm{f}(x)=x^2-2x-3$, the curve $y=x^2-2x+1=\mathrm{f}(x)+4$ is a translation by $\begin{pmatrix}0\\4\end{pmatrix}$, and $y=\frac{x^2}{4}-x-3=\mathrm{f}\left(\frac x2\right)$ is a stretch parallel to the $x$-axis with scale factor $2$.