Transformations of Graphs

Two places a number can act

Start with one concrete case. Suppose $(2, 5)$ lies on the graph of $y = f(x)$, and you are told $y = f(x - 3) + 1$. The old input $2$ reappears when the inside equals it, so solve $x - 3 = 2$ to get the new input $x = 5$. The old height $5$ becomes $5 + 1 = 6$ once the outside adds to it. So this rule sends the point to $(5, 6)$: the $-3$ changed which input you asked about, and the $+1$ changed only the height once you already had an answer. That is the whole lesson in miniature, one number acting on the question you ask $f$, another acting on the answer it gives back.

Write a general transformed rule in the shape you will unpack for the rest of the lesson:

There are exactly two places a number can sit, and they behave completely differently.

The numbers $a$ and $k$ act on the output, the value $f$ has already produced. You take the answer $f(\cdot)$, multiply it by $a$, then add $k$. These are the moves that behave the way you expect: multiply the height, then raise it. Call this the outside, the part applied after $f$.

The numbers $b$ and $h$ act on the input, the number handed to $f$ before $f$ does anything. The rule first forms $x - h$, then multiplies by $b$, and only then feeds the result to $f$. These are the moves that make the graph go the opposite way from what the symbols seem to say. Call this the inside, the part applied before $f$.

Each side applies one step to a number: on the inside you change $x$ before handing it to $f$, and on the outside you change the number $f$ hands back. Writing $f(g(x))$ for “do $g$, then do $f$” is the same substitution you practiced in Function Notation, so here we only ever say “apply this, then $f$, then that,” one step at a time. The whole lesson is learning to read those two sides correctly.

Why the inside runs backward

Start with the inside, because it is where the confusion lives. Take a pure horizontal shift, $g(x) = f(x - h)$, and ask not what the formula looks like but where the points actually go.

Because this is the move most often read backward, fix the vocabulary now and keep it fixed. The inside factor $b$ divides every input. When $b > 1$ that pulls the graph toward the vertical axis, and we call it a horizontal compression by $b$. When $0 < b < 1$ the inputs grow instead and the graph is stretched away from the vertical axis. Throughout this lesson the number named in a compression is always the input multiplier $b$, never its reciprocal. So “a horizontal compression by $b$” always means “the inputs are divided by $b$.”

Here is the one idea that explains the whole inside, and it is worth saying out loud. An outside map is applied to the answer $f$ already gave, so it does exactly what it says: multiply the answer, add to the answer. An inside map is applied to the question, the number you are about to hand $f$. So to find where a point lands you must ask which $x$ makes the inside equal the old input, and answering that means solving for $x$. The solving is what turns “subtract $h$” into “move right by $h$” and “multiply by $b$” into “divide by $b$.”

The master formula

Both sides fold into a single statement about where a point goes. It is the tool the rest of the lesson uses, so derive it once and keep it.

Every rule from the earlier course is one line of this formula. Setting $a = b = 1$ and $h = 0$ gives $(u,\, v + k)$, a vertical shift by $k$. Setting $a = b = 1$ and $k = 0$ gives $(u + h,\, v)$, a horizontal shift right by $h$. Setting $b = 1$, $h = 0$, $k = 0$ gives $(u,\, a v)$, a vertical scaling by $a$, which also reflects across the $x$-axis when $a < 0$. Setting $a = 1$, $h = 0$, $k = 0$ gives $\left(\tfrac{u}{b},\, v\right)$, which divides every input position by $b$ (a horizontal compression when $b > 1$) and reflects across the $y$-axis when $b < 0$. The one formula carries all of them, and the two halves never mix. The input data $u$ is touched only by $b$ and $h$, and the output data $v$ only by $a$ and $k$.

That separation reaches the domain and range too, the data Relations and Functions insisted was part of a function. The inputs $u$ of $f$ become $\tfrac{u}{b} + h$, so the domain is divided by $b$ and shifted by $h$. The outputs $v$ become $a v + k$, so the range is scaled by $a$ and shifted by $k$. In general the transformed domain is the set of every value $\tfrac{u}{b} + h$ for $u$ in the original domain, and the transformed range is the set of every value $av + k$ for $v$ in the original range, the same rule that moves every other point on the graph. When the domain or range is a plain interval, a negative $a$ reverses the order of the range’s endpoints, since it turns the larger output into the smaller one, and a negative $b$ does the same to the domain’s endpoints.

A domain is easier to trust once you have watched one move. The graph below is $y = a\sqrt{x - h} + k$, and a square root is the useful case because its domain has a visible left end rather than running off both sides of the window. Drag $h$ upward and watch the start point: it travels right, against the minus sign inside, and the whole graph goes with it, so the domain becomes $x \ge h$ and the readout confirms it. That is the same solve-the-inside reasoning as the worked examples above, now watched instead of computed.

Two states are worth reaching deliberately. Drag $a$ negative and watch the curve turn over to fall instead of rise, while its left end does not move at all: the left end stays put because $a$ acts outside and the domain is decided inside. Drag $a$ to zero and watch the graph flatten onto a horizontal line, and yet the readout still reports the domain restriction, because multiplying by zero cannot widen what the root will accept.

Order matters, and the two sides disagree

The master formula hides a subtlety that causes more errors than any sign. When a side carries two operations, the order in which you apply them to the graph is not free.

On the outside, the graph order matches the arithmetic order. The rule $y = a f(x) + k$ says compute $f(x)$, then multiply by $a$, then add $k$. As graph moves that reads scale by $a$ first, then shift by $k$, in that order, because the arithmetic is done in that order to every height.

On the inside, the graph order is the reverse of the arithmetic order, and the reason is the solving step. The rule $y = f\big(b(x - h)\big)$ processes $x$ by first subtracting $h$, then multiplying by $b$, then handing the result to $f$. But the master formula shows the point lands at $\tfrac{u}{b} + h$, which you build from $u$ by first dividing by $b$ (the horizontal scaling) and then adding $h$ (the shift right). Scaling comes first, shift second, the opposite of the arithmetic order. That reversal is forced: to solve $b(x - h) = u$ for $x$ you undo the multiplication before you undo the subtraction, and undoing in reverse is what solving an equation always does.

This reversal is exactly why the inside must be factored into the form $b(x - h)$ before you can read the shift. Skipping that step produces the single most common error in all of this material.

When the order of two moves changes the result

The order rule has a sharp consequence: doing two moves in one order can give a genuinely different function from doing them in the other order.

Take a vertical stretch by $2$ and a vertical shift up $3$, and apply them to $y = f(x)$ in each order. Stretch first, then shift, and you get $2f(x)$ and then $2f(x) + 3$. Shift first, then stretch, and you get $f(x) + 3$ and then $2\big(f(x) + 3\big) = 2f(x) + 6$. These are not the same rule:

The gap is the extra $3$ that the second stretch also multiplied. So “stretch then shift” and “shift then stretch” are two different transformations, and the graphs land in different places.

The same reversal happens when both moves act on the input instead of the output. Reflect $f$ across the $y$-axis and then shift right $2$, and you build $f\big(-(x - 2)\big) = f(2 - x)$: the shift is applied to the already-reflected rule, so it lands inside the negation. Shift first and reflect second, and the $2$ never gets negated: you build $f(-x - 2)$ instead. Two input-side moves disagree about order exactly as two output-side moves do, and for the same reason, each new move is applied to whatever rule the previous move already produced.

One graph, many transformations

It is tempting to believe the finished graph tells you which transformation produced it. It does not, and seeing why is one of the honest surprises of this lesson.

Take $f(x) = x^2$ and compress it horizontally by $2$:

Now instead stretch the same $f$ vertically by $4$:

They are the same function, $y = 4x^2$, reached by two different transformations. So you may truthfully say ”$y = 4x^2$ is $y = x^2$ compressed horizontally by $2$,” and equally truthfully say “it is $y = x^2$ stretched vertically by $4$.” What you may not say is that the transformation is unique, or that the graph reveals which move was applied. Both descriptions are correct, and the graph cannot choose between them.

This is not special to $x^2$: for positive $b$, every pure power function $f(x) = x^d$ satisfies $f(bx) = (bx)^d = b^d x^d = b^d f(x)$, so a horizontal compression by $b$ can always be redescribed as an outside scaling by $b^d$ (for example, $f(2x) = 8f(x)$ when $f(x) = x^3$). Pure power functions are one important family with this property, and adding a constant already breaks it. For $f(x) = x^2 + 1$,

can never equal $4x^2 + 1$, since matching the $x^2$ term needs $c = 4$ while matching the constant needs $c = 1$. So here the horizontal compression $f(2x)$ is not any vertical stretch at all, and the two kinds of scaling are genuinely different operations. The figure shows this second case, where $f(2x)$ and $2f(x)$ are plainly different curves.

Symmetry is a transformation that does nothing

The chapter’s earlier lesson defined a function as even when $f(-x) = f(x)$ and odd when $f(-x) = -f(x)$, both requiring a domain symmetric about $0$ so that $-x$ is available whenever $x$ is. Read through the lens of this lesson, those definitions are statements about a single inside transformation, reflecting the input, written $y = f(-x)$.

An even function is one that reflecting the input leaves completely unchanged: $f(-x) = f(x)$ says the reflected graph $y = f(-x)$ is the very same graph as $y = f(x)$. So evenness is precisely invariance of the graph under reflection across the $y$-axis. An odd function is one where reflecting the input has the same effect as reflecting the output: $f(-x) = -f(x)$ says the inside reflection $y = f(-x)$ equals the outside reflection $y = -f(x)$. Doing both at once, reflecting input and output together, is a $180^\circ$ rotation about the origin, and it returns the original graph, since $-f(-x) = f(x)$. So oddness is invariance of the graph under that rotation.

Both statements need the symmetric domain to even be asked, exactly as the earlier lesson proved, and neither says more than it should. In particular an odd function need not pass through the origin: it does when $0$ is in its domain, because then $f(0) = -f(0)$ forces $f(0) = 0$. But $f(x) = \tfrac{1}{x}$ is odd with $0$ absent from its domain, and it never meets the origin. Symmetry, in this chapter’s language, is just the special case where a particular transformation fixes the graph.

Common mistakes