Combinations of transformationsAQA A-Level Maths: Revision notes

Several transformations can act on the same graph, as in $y=af(x+b)+c$. The final graph depends on the order in which they are applied, so work through them in the correct sequence.

• Changes inside the brackets, $f(x+b)$ and $f(ax)$, affect $x$-coordinates, and they act in the opposite sense to what you might expect.
• Changes outside, $af(x)$ and $+c$, affect $y$-coordinates and act as written. Inside and outside changes do not interfere with one another, so you can treat the $x$-changes and $y$-changes separately.

Within one direction the order matters. For $y=af(x)+c$, the stretch comes first and then the translation, because that is the order you would evaluate it in: multiply $f(x)$ by $a$, then add $c$. Example: $y=2x^2+1$ is $y=x^2$ stretched by factor $2$ in $y$, then moved up $1$. Doing the translation first would give $y=2(x^2+1)=2x^2+2$ instead. For $x$: $y=f(2x-4)=f\big(2(x-2)\big)$, so the graph is stretched by factor $\frac12$ in $x$ first, and then translated $2$ to the right.

A point $(x,y)$ on $y=f(x)$ becomes $(x-b,\ ay+c)$ on $y=af(x+b)+c$. Example: $y=f(x)$ has a minimum at $(1,-3)$. On $y=2f(x)+1$ the minimum is $(1,2(-3)+1)=(1,-5)$. On $y=f(2x-4)$, solve $2x-4=1$: the minimum is at $x=\frac52$, so $\left(\frac52,-3\right)$. In general, for $y=f(ax+b)$ the image of $x_0$ is $\frac{x_0-b}{a}$. Always track every labelled point: turning points, intercepts and asymptotes.

A negative multiplier, as in $y=-f(x)$ or $y=3-f(x)$, includes a reflection in the $x$-axis. Reflect first, then translate: $3-f(x)=-f(x)+3$. A minimum at $(1,-3)$ on $y=f(x)$ becomes a maximum at $(1,6)$ on $y=3-f(x)$. For $y=-g(2x)$ with a maximum of $g$ at $(-2,5)$: the $x$-coordinate halves to $-1$, the $y$-coordinate changes sign to $-5$, and the maximum becomes a minimum, giving a minimum at $(-1,-5)$.

To describe how $y=x^2$ becomes $y=2(x-3)^2+1$, give each step in order:

1. translation $3$ units right: $y=(x-3)^2$;
2. stretch with scale factor $2$ parallel to the $y$-axis: $y=2(x-3)^2$;
3. translation $1$ unit up: $y=2(x-3)^2+1$. The vertex moves from $(0,0)$ to $(3,1)$, and the $y$-intercept is $2(9)+1=19$. Since the minimum value is $1>0$, the curve never meets the $x$-axis. Given the images of two points you can find unknown constants. If $(2,6)\to(-1,10)$ and $(5,0)\to(2,-2)$ under $y=af(x+b)+c$: $2-b=-1$ gives $b=3$; $c=-2$ from the point on the axis; $6a-2=10$ gives $a=2$.