Two changes at once, and why the order matters

Combining transformations, reading them off a formula, and the factoring trap.

Putting on socks and then shoes gives a different result from putting on shoes and then socks. Transformations are the same: doing two of them in the other order usually lands the graph somewhere else.

Here are the same two changes applied in the two possible orders. Run it and compare the columns.

At $x = 0$ the first column gives 3 and the second gives 6. Stretching first and then raising by 3 lifts the graph by 3. Raising by 3 and then stretching lifts it by 3 and then doubles that lift as well.

So the order is not a matter of taste. It is decided by the formula, and reading it correctly is reading the brackets correctly.

The general form

Every combination of the four transformations can be written in one shape.

Work from the inside out, because that is the order the arithmetic actually happens in. The input meets $x - h$ first, then the multiplication by $b$, then $f$ itself, then the multiplication by $a$, then the addition of $k$.

1. Horizontal: shift by $h$ and squash by $b$. These two act on the input.
2. Then apply the parent function.
3. Vertical: stretch by $a$, then shift by $k$. These two act on the output, stretch before shift.

The vertical pair is where marks are lost. In $a f(x) + k$ the stretch is applied to $f(x)$ and not to $k$, so the shift happens after the stretch. Someone who shifts first and then stretches moves the graph $ak$ instead of $k$.

Before moving the sliders, fix what $b$ does. The widget below opens with $b = 2$, which means the height the plain parabola reached one unit from its centre is now reached half a unit from the centre. Drag $b$ down to 0.5 and the opposite happens: what was reached at one unit from the centre is now reached at two units, so the curve widens.

The trap: factor the inside first

A formula does not always arrive in the tidy form. Consider $y = (2x - 6)^2$. It is tempting to read the 6 as a shift of 6 to the right, and that is wrong. Check it against a table.

The output is zero at $x = 3$, not at $x = 6$. Factoring the inside shows why: $2x - 6 = 2(x - 3)$, so $y = \bigl(2(x-3)\bigr)^2$, which is a shift of 3 right and a horizontal squash by a factor of 2.

One worked example in full. Describe the transformations that take $y = x^2$ to $y = -3(2x + 8)^2 + 1$, in order.

1. Factor the inside: $2x + 8 = 2(x + 4)$, so the rule is $y = -3\bigl(2(x+4)\bigr)^2 + 1$.
2. Horizontal shift: $x + 4$ is $x - (-4)$, so $h = -4$, a move of 4 to the left.
3. Horizontal squash: $b = 2$, so every horizontal distance from the centre is halved.
4. Vertical stretch: $a = -3$, so heights are tripled and then flipped over the horizontal axis.
5. Vertical shift: $k = 1$, so finally the whole thing rises by 1.
6. Check with a point: at $x = -4$ the output is $-3(0) + 1 = 1$, so the turning point is $(-4, 1)$, which agrees.