Transformations of functions

In $g(x) = af(b(x - h)) + k$, the constants $a$ and $k$ sit outside the function. They act on outputs, after $f$ has done its work, and they behave literally: $k$ slides the graph up by $k$ (down if negative), and $a$ stretches every output by a factor of $|a|$, flipping the graph across the x-axis if $a$ is negative.

The constants $b$ and $h$ sit inside. They act on inputs, before $f$ ever runs, and they behave in reverse: $x - h$ slides the graph right by $h$, so $f(x + 3)$ moves left 3. A factor $b$ compresses the graph horizontally by $1/|b|$, so $f(2x)$ is squeezed to half its width, not doubled. If $b$ is negative, the graph reflects across the y-axis.

One sentence carries the whole topic: outside changes do what they say to outputs; inside changes do the opposite to inputs.

The reversal is not a convention to memorize; it falls out of asking what input the new function needs. Take $g(x) = f(x - 3)$. For $g$ to produce the value $f$ used to produce at 0, it needs $x - 3 = 0$, that is, $x = 3$. Every feature of $f$ happens 3 units later on the x-axis: the graph moved right, even though the formula says minus.

Same logic for $g(x) = f(2x)$. To reproduce $f(6)$, the new function needs $2x = 6$, so $x = 3$. What used to happen at 6 now happens at 3: everything pulled toward the y-axis, a compression, even though the formula multiplies.

When in doubt, do not recall a rule. Set the inside expression equal to the old input and solve. The algebra always tells the truth about direction.

The form $af(b(x - h)) + k$ only reads correctly when the inside is factored. Given $g(x) = f(2x + 6)$, the shift is not 6. Factor first: $f(2x + 6) = f(2(x + 3))$. Now the dials are visible: horizontal compression by $1/2$, then translation left 3.

Skipping the factoring step is the single most common transformation error on the exam. Any time $b \neq 1$ and a constant rides along inside, factor $b$ out before saying a single word about shifts.

The fastest way to check any transformation is to push one known point through it. If $(p, q)$ is on $f$, then on $g(x) = af(b(x - h)) + k$ the point lands at $\left(\frac{p}{b} + h,\; aq + k\right)$: input operations (divide by $b$, then shift by $h$) on the x-coordinate, output operations ($\times a$, then $+k$) on the y-coordinate.

Domain and range follow the same split. Inside changes act on the domain: $f(x - 5)$ slides the whole domain right 5, and $f(2x)$ halves every domain endpoint. Outside changes act on the range: $3f(x)$ triples the range endpoints, $f(x) + 2$ raises them by 2. A translation slides an interval without changing its length; a dilation rescales it.

Order matters on the output side: in $2f(x) + 3$ the stretch happens before the shift, because that is the order the arithmetic applies to each output value. Reading $-f(x) + 2$ as "shift up 2, then flip" produces the wrong graph; the flip comes first.