Transformations of graphsEdexcel International A Level Maths: Revision notes

$y=f(x)+a$ is a translation of $a$ units up: every $y$-coordinate increases by $a$. $y=f(x+a)$ is a translation of $a$ units to the left: the change inside the bracket acts on $x$ and works in the opposite direction to the sign. So $y=f(x-3)$ moves the graph $3$ to the right. Example: if $f$ has a minimum at $(2,-3)$, then $f(x)+4$ has a minimum at $(2,1)$ and $f(x-3)$ has a minimum at $(5,-3)$.

$y=af(x)$ is a stretch parallel to the $y$-axis with scale factor $a$: every $y$-coordinate is multiplied by $a$. If $a<0$ it also reflects in the $x$-axis, so $y=-f(x)$ is a reflection in the $x$-axis. $y=f(ax)$ is a stretch parallel to the $x$-axis with scale factor $\frac1a$: every $x$-coordinate is divided by $a$. So $y=f(2x)$ halves the $x$-coordinates, and $y=\sin2x$ has period $180^\circ$. Points on the axis the graph is stretched away from stay put: the roots do not move in a vertical stretch.

Apply the rules to the key points of the graph. For $y=x^2$: $y=(x-2)^2+1$ has its vertex at $(2,1)$. For $y=\frac1x$: $y=\frac1{x+2}-3$ has asymptotes $x=-2$ and $y=-3$, and crosses the axes at $\left(-\frac53,0\right)$ and $\left(0,-\frac52\right)$. For trigonometric graphs: $y=\sin(x-30^\circ)$ has its first maximum at $(120^\circ,1)$; $y=\cos x+2$ has maximum $3$ and minimum $1$; $y=3\sin x$ has maximum $3$; $y=\tan2x$ has asymptotes at $45^\circ$, $135^\circ$, ... Cubics work the same way: $y=(x-1)^3$ is $y=x^3$ translated $1$ to the right.

Given the graph of $y=f(x)$, sketch the new graph by transforming key points: intercepts, turning points and asymptotes. Label the new coordinates. Example: with $f(x)=x^2-4x+1=(x-2)^2-3$, $y=f(x+3)=x^2+2x-2$ has its minimum at $(-1,-3)$; $y=f(2x)=4x^2-8x+1$ has its minimum at $(1,-3)$; $y=-f(x)$ has its maximum at $(2,3)$. Apply one transformation at a time, and check by substituting a point.