Ask: is the change inside or outside the function bracket?

• Outside $f(x)$ — forms $f(x)+a$ and $af(x)$ — affects the $y$-coordinate and behaves intuitively (vertical).
• Inside $f(x)$ — forms $f(x+a)$ and $f(ax)$ — affects the $x$-coordinate and behaves counter-intuitively (horizontal).

Identifying which family you're in before trying to recall the specific rule prevents almost every common error.

$y=f(x)+a$ translates the graph $a$ units up (point mapping: $(x,y)\to(x,y+a)$).

$y=af(x)$ stretches vertically by scale factor $a$ (mapping: $(x,y)\to(x,ay)$); if $a=-1$, this is a reflection in the $x$-axis.

$y=f(x+a)$ shifts $a$ units left (not right) — mapping $(x,y)\to(x-a,y)$.

$y=f(ax)$ compresses horizontally by scale factor $\frac{1}{a}$ — mapping $(x,y)\to(\frac{x}{a},y)$; if $a=-1$, this reflects in the $y$-axis.

Why the reversal? Think of it as "coordinate compensation." If $f(10)=5$, then for $g(x)=f(x+3)$ to also output 5, you need $x+3=10$, i.e. $x=7$ — three less than before. The point moved from $(10,5)$ to $(7,5)$: three units left. The $+3$ forces $x$ to shrink to compensate.

Transformations don't commute — order matters. Apply stretches before shifts (like BIDMAS: multiplication before addition).

For the trickiest form, $y=f(ax+b)$, you must factorise before reading off the shift: rewrite as $f(a(x+\frac{b}{a}))$. For example, $y=f(2x+6) = f(2(x+3))$ means: shift 3 left, then compress by $\frac{1}{2}$ — the true horizontal shift is 3 units, not 6.

Writing $y=ax^2+bx+c$ as $y=a(x-h)^2+k$ shows the full transformation chain from the parent $y=x^2$: horizontal shift $h$, vertical stretch $a$, vertical shift $k$. The vertex of $y=x^2$ at $(0,0)$ maps to $(h,k)$ — which is exactly why $(h,k)$ is the vertex of any quadratic in this form.

Worked example: Describe the transformation mapping $y=x^2$ onto $y=x^2+6x+10$.

This is $f(x+3)+1$: translation by $\begin{pmatrix}-3\\1\end{pmatrix}$ (3 left, 1 up).

For $y=a\sin(bx)+c$: amplitude becomes $|a|$, principal axis moves to $y=c$, and period becomes $\frac{360°}{b}$.

For a reciprocal graph like $y=\frac{1}{x-3}+4$ (i.e. $f(x-3)+4$ where $f(x)=\frac{1}{x}$): the vertical asymptote $x=0$ shifts to $x=3$, and the horizontal asymptote $y=0$ shifts to $y=4$. Asymptotes transform exactly like any other feature of the graph.

The minimum point of $y=f(x)$ is at $P(3,-4)$. Find the coordinates of the minimum point of $y=f(x+5)+6$.

Identify: $a=+5$ inside the bracket → horizontal shift 5 left. The $+6$ outside → vertical shift 6 up.

Apply the rule $(x,y)\to(x-5,y+6)$ to $P(3,-4)$: