Transformations of Graphs

Intuition

In an earlier course you learned a stack of transformation rules and where each one moves a graph. Adding outside slides it up, subtracting inside slides it right, a factor bigger than $1$ inside squeezes it narrower, and the inside changes always look backward. You met all of that as facts to memorize. This lesson does not add new rules; it reveals that every one of them is the same rule. A transformation is a map applied in exactly one of two places, to the number you feed $f$ before $f$ acts, or to the answer $f$ returns after it acts. That single split explains every sign, every direction, and every ordering mistake in this lesson, including the two that trip up almost everyone. Those two are why $f(2x - 3)$ does not shift by $3$ , and why the same finished graph can come from more than one transformation. Once you see the split, the rules stop being a list and become one idea.

Learning goals

Separate outside moves on the output from inside moves on the input
Explain why the inside runs backward, by solving for $x$
Apply the master formula to send $(u, v)$ to its image
Factor the inside first, so $f(2x - 3)$ shifts by $\tfrac{3}{2}$
Show that two moves need not commute
Read symmetry as a transformation that changes nothing

Two places a number can act

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

y = a\,f\big(b(x - h)\big) + k.

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 is a small chain of maps: on the inside you apply one map to $x$ and then hand the result to $f$ . On the outside you hand $f$ ‘s answer to another map. Writing $f(g(x))$ for “do $g$ , then do $f$ ” is just the substitution you already practiced in Function Notation. Combining functions this way is the subject of the next lesson. So here we only ever say “apply this, then $f$ , then that,” and never lean on any special algebra of that combining. 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.

An inside map moves the graph opposite to its sign#

Let $g(x) = f(x - h)$ . A point $(x, y)$ lies on the graph of $g$ exactly when $y = g(x) = f(x - h)$ , and that says precisely that the point $(x - h, y)$ lies on the graph of $f$ . Read that correspondence starting from $f$ . Take a point sitting at horizontal position $u$ on the graph of $f$ , namely $(u, f(u))$ . It shows up on the graph of $g$ at the input $x$ where the inside equals $u$ , that is where $x - h = u$ , so $x = u + h$ . Its height $f(u)$ is unchanged. Every point therefore moves from position $u$ to position $u + h$ , a slide to the right by $h$ when $h > 0$ , even though the formula shows a minus sign.

The same reading handles a pure scaling, $g(x) = f(bx)$ with $b > 0$ . Here $(x, y)$ is on $g$ exactly when $y = f(bx)$ , which says $(bx, y)$ is on $f$ . A point at position $u$ on $f$ reappears on $g$ where $bx = u$ , that is at $x = \dfrac{u}{b}$ , again at the same height. So the horizontal position of every point is divided by $b$ , which pulls points toward the vertical axis when $b > 1$ and pushes them away when $0 < b < 1$ .

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 $b$ .” (That step, solving the inside for $x$ , is the seed of the inverse function you will meet two lessons from now.)

The dashed arch is y = f(x); the solid arch is y = f(x - 3). Reading from f to its shifted copy, the peak that sat at input 2 reappears at input 5, because the inside x - 3 equals 2 exactly when x = 5. Every point moves right by 3, even though the formula shows a minus sign.

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.

Where every point of the graph lands#

Fix the general rule $y = a\,f\big(b(x - h)\big) + k$ with $b \ne 0$ , and take any point $(u, v)$ on the graph of $f$ , so $v = f(u)$ . We find the point $(x, y)$ it becomes on the new graph.

The new rule reproduces the height $v = f(u)$ when the number handed to $f$ equals $u$ , that is when $b(x - h) = u$ . This is an equation in $x$ , and solving it means dividing by $b$ and then adding $h$ :

b(x - h) = u \ \Rightarrow\ x - h = \frac{u}{b} \ \Rightarrow\ x = \frac{u}{b} + h.

At that input the output is $a\,f(u) + k = a v + k$ , read straight off the outside. So the point $(u, v)$ on the graph of $f$ lands at

(u, v) \ \longmapsto\ \left(\frac{u}{b} + h,\ \, a v + k\right).

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 $k$ . A negative $a$ or $b$ flips the corresponding interval end for end. The domain and range transform right along with the picture, because they are the picture.

Worked example 1 Reading the master formula

The point $(4, 6)$ lies on the graph of $f$ . Where does it land on $y = 2f(x - 3) + 1$ , and on $y = f(2x)$ ?

For $y = 2f(x - 3) + 1$ , match the shape $a\,f\big(b(x - h)\big) + k$ : here $a = 2$ , $b = 1$ , $h = 3$ , $k = 1$ . Apply the landing rule $\left(\tfrac{u}{b} + h,\ a v + k\right)$ to $(u, v) = (4, 6)$ :

\left(\frac{4}{1} + 3,\ \, 2\cdot 6 + 1\right) = (7, 13).

The input picked up the shift $+3$ and the height was doubled and then raised by $1$ .

For $y = f(2x)$ , the outside is untouched ( $a = 1$ , $k = 0$ ) and the inside has $b = 2$ , $h = 0$ . The landing rule gives

\left(\frac{4}{2} + 0,\ \, 6\right) = (2, 6).

The height is unchanged and the input is halved, so the point slides in toward the vertical axis. Both results come from the same formula, just with different slots filled.

Worked example 2 Transforming a domain and a range

Suppose $f$ has domain $2 \le x \le 8$ and range $-1 \le y \le 5$ . Describe the domain and range of $y = f(2x)$ and of $y = -3f(x)$ .

For $y = f(2x)$ , only the inside changes, with $b = 2$ . The domain is divided by $b = 2$ , so each endpoint is halved:

2 \le x \le 8 \ \longrightarrow\ 1 \le x \le 4.

The range is untouched, still $-1 \le y \le 5$ , because nothing acted on the output.

For $y = -3f(x)$ , only the outside changes, with $a = -3$ . The range is scaled by $a = -3$ , and because $a$ is negative the endpoints swap order:

-3 \cdot (-1) = 3, \qquad -3 \cdot 5 = -15,

so the new range runs from the smaller value up, $-15 \le y \le 3$ . The reflection is exactly what turned the top of the old range into the bottom of the new one. The domain stays $2 \le x \le 8$ , since the inside was left alone.

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. Step $h$ upward: the start point travels right, against the minus sign inside, and the whole graph goes with it, so the domain becomes $x \ge h$ and the readout says so. That is the same halving-and-shifting arithmetic as the worked example above, done by hand.

Two states are worth reaching deliberately. Take $a$ negative and the curve turns 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. Take $a$ to zero and the graph flattens onto a horizontal line, and yet the readout still reports the domain restriction. Multiplying by zero cannot widen what the root will accept.

Move the start point of $y = a\sqrt{x - h} + k$

Stretch a Inside h Outside k

y = √x. Its start point sits at (0, 0), not shifted at all. From there it rises to the right, at the natural rate, flattening as it goes. It exists only for x of at least 0, because a square root needs the inside to be at least 0.

A square root graph on a coordinate plane. The inside shift moves both the start point and the left end of the domain, while the coefficient outside changes only how steeply the curve leaves that point and which way it goes.

Check your understanding

The point $(6, 2)$ lies on the graph of $y = f(x)$ . Where does it land on the graph of $y = 2f(x - 1) + 3$ ?

(5, 7)

(7, 7)

(7, 4)

(12, 3)

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.

Worked example 3 The factoring trap: reading f(2x - 3)

The rule $y = f(2x - 3)$ is not ” $f(2x)$ shifted right $3$ .” To read it correctly, factor the inside so it matches the shape $b(x - h)$ :

2x - 3 = 2\left(x - \tfrac{3}{2}\right).

Now the slots are visible: $b = 2$ and $h = \tfrac{3}{2}$ . So the graph is a horizontal compression by $2$ (the inputs are divided by $2$ ) and a shift right by $\tfrac{3}{2}$ , not by $3$ .

Check it against the master formula, which never needs the factoring. A point $(u, v)$ on $f$ lands where $2x - 3 = u$ , that is at

x = \frac{u + 3}{2} = \frac{u}{2} + \frac{3}{2}.

The shift added to the scaled input is $\tfrac{3}{2}$ , matching the factored reading. The tempting wrong answer, “shift right $3$ ,” comes from reading the $-3$ before dividing out the $2$ , which reverses the required order of the two inside moves. The inside multiplier $2$ has to be pulled out first.

Check your understanding

After its horizontal compression, by how much and which way does $y = f(3x - 12)$ shift the graph of $y = f(x)$ ?

A right

12

B left

4

C right

4

D right

\tfrac{12}{3} = 4

then negated to left

4

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. This is your first real meeting with the fact that composing maps does not commute, which is the headline of the next lesson.

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:

2f(x) + 3 \ \ne\ 2f(x) + 6.

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.

Both curves come from stretching y = x squared by 2 and shifting up 3, in opposite orders. Stretch then shift gives y = 2f(x) + 3 with vertex (0, 3); shift then stretch gives y = 2(f(x) + 3) = 2f(x) + 6 with vertex (0, 6). Same stretch, same shift, different order, and the graphs sit 3 apart.

Worked example 4 A reflection and a shift, in both orders

Reflection across the $x$ -axis (the outside factor $a = -1$ ) also refuses to commute with a shift. Track the point $(2, 3)$ on $f$ through both orders.

Reflect first, then shift up $5$ . Reflecting negates the height, $(2, 3) \to (2, -3)$ , then shifting adds $5$ :

(2, -3) \ \longrightarrow\ (2,\, -3 + 5) = (2, 2).

As a rule this is $y = -f(x) + 5$ . Now shift up $5$ first, then reflect. Shifting gives $(2, 3) \to (2, 8)$ , then reflecting negates that height:

(2, 8) \ \longrightarrow\ (2,\, -8).

As a rule this is $y = -\big(f(x) + 5\big) = -f(x) - 5$ . The two landing points, $(2, 2)$ and $(2, -8)$ , are different, and the two rules, $-f(x) + 5$ and $-f(x) - 5$ , are different. The reflection multiplies by $-1$ everything that is already there, so whether the $+5$ is present when you reflect changes the result. Order is part of the transformation, not a detail you can rearrange.

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 $2$ :

f(2x) = (2x)^2 = 4x^2.

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

4f(x) = 4x^2.

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 coincidence is special to homogeneous power functions, not to $x^2$ in particular, and it is not the typical case. It happens because a power pulls the factor straight through: for $f(x) = x^d$ , $f(bx) = (bx)^d = b^d x^d = b^d f(x)$ , so a horizontal compression by $b$ is the same as a vertical stretch by $b^d$ . The cube behaves the same way, with $f(2x) = 8f(x)$ for $f(x) = x^3$ . What kills the coincidence is breaking the homogeneity by adding a constant. For $f(x) = x^2 + 1$ ,

f(2x) = 4x^2 + 1, \qquad \text{but} \qquad c\,f(x) = c\,x^2 + c

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.

For f(x) = x squared + 1 the two kinds of scaling differ. The narrow curve is y = f(2x) = 4x squared + 1, a horizontal compression, with vertex (0, 1). The wide curve is y = 2f(x) = 2x squared + 2, a vertical stretch, with vertex (0, 2). They are different graphs, and 4x squared + 1 is not any vertical stretch of f, so the coincidence seen for x squared has vanished.

Worked example 5 Same graph, two honest descriptions

Show that $y = 9x^2$ can be obtained from $y = x^2$ as either a horizontal compression or a vertical stretch, and give the exact number for each.

As a horizontal compression, replace $x$ by $bx$ and ask which $b$ gives $9x^2$ :

f(bx) = (bx)^2 = b^2 x^2 = 9x^2 \ \Rightarrow\ b^2 = 9 \ \Rightarrow\ b = 3,

a horizontal compression by $b = 3$ , since the inputs are divided by $3$ . As a vertical stretch, multiply the output by $a$ :

a\,f(x) = a\,x^2 = 9x^2 \ \Rightarrow\ a = 9,

a stretch to $9$ times the height. Both produce the identical graph $y = 9x^2$ , so the correct statement is that the transformation is not unique. That transformation can be described as a horizontal compression by $3$ or as a vertical stretch by $9$ . The graph itself does not record which one you had in mind. Try the same on $g(x) = x^2 + 1$ and the vertical route fails, since $g(3x) = 9x^2 + 1$ is not $a(x^2 + 1)$ for any $a$ .

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.

Check your understanding

A function $f$ is odd, with a domain symmetric about $0$ . Which single transformation is guaranteed to leave its graph unchanged?

A reflection across the

x

-axis B reflection across the

y

-axis C a

180^\circ

rotation about the origin D a vertical shift up

1

Common mistakes

Reading a horizontal shift without factoring. $y = f(2x - 3)$ is not ” $f(2x)$ shifted right $3$ .” Factor the inside as $2\left(x - \tfrac{3}{2}\right)$ : the shift is right $\tfrac{3}{2}$ , because the factor $b$ must be pulled out before the shift $h$ is visible.
Applying the inside moves in arithmetic order. On the inside the graph order is the reverse of the arithmetic order, since finding where a point lands means solving the inside for $x$ . Scale first, then shift, the opposite of “subtract then multiply.”
Assuming two moves commute. $2f(x) + 3$ (stretch then shift) and $2(f(x) + 3) = 2f(x) + 6$ (shift then stretch) are different functions. Reordering a stretch and a shift, or a reflection and a shift, generally changes the result.
Believing the graph names its transformation. For $f(x) = x^2$ , the graph $y = 4x^2$ is both a horizontal compression by $2$ and a vertical stretch by $4$ . The finished graph does not reveal which was applied, so never claim the transformation is unique.
Moving the wrong coordinate. An outside change ( $a$ , $k$ ) touches only the height, so $(u, v) \to (u, av + k)$ ; an inside change ( $b$ , $h$ ) touches only the input, so $(u, v) \to \left(\tfrac{u}{b} + h,\ v\right)$ . Do not shift or scale both coordinates for a one-sided change.
Testing even or odd on an unsymmetric domain. Both $f(-x) = f(x)$ and $f(-x) = -f(x)$ require that $-x$ be in the domain whenever $x$ is; if the domain is not symmetric about $0$ , the question cannot be asked.
Claiming every odd function passes through the origin. That holds only when $0$ is in the domain; $f(x) = \tfrac{1}{x}$ is odd yet never meets the origin.

Key takeaways

Every transformation acts in one of two places: the outside ( $a$ , $k$ ) on the output after $f$ , or the inside ( $b$ , $h$ ) on the input before $f$ . The outside behaves as written; the inside runs backward.
The inside runs backward because it acts on the number you hand $f$ , so finding where a point lands means solving the inside for $x$ . That is why $y = f(x - h)$ moves right and $y = f(bx)$ divides every input by $b$ (a horizontal compression by $b$ when $b > 1$ ).
Master formula: under $y = a\,f\big(b(x - h)\big) + k$ , a point $(u, v)$ on $f$ lands at $\left(\tfrac{u}{b} + h,\ a v + k\right)$ . The domain is divided by $b$ and shifts by $h$ ; the range scales by $a$ and shifts by $k$ .
Order matters, oppositely on the two sides. Outside moves apply in arithmetic order (scale then shift); inside moves apply in reverse (scale then shift for $b(x - h)$ , the opposite of the arithmetic “subtract then multiply”).
Factor the inside first. $f(2x - 3) = f\left(2\left(x - \tfrac{3}{2}\right)\right)$ is a compression by $2$ and a shift right $\tfrac{3}{2}$ , not $3$ .
Two moves need not commute: $2f(x) + 3 \ne 2(f(x) + 3)$ . This is the non-commuting of composition that the next lesson studies.
A graph does not determine its transformation. For $f(x) = x^2$ , $y = 4x^2$ is both $f(2x)$ and $4f(x)$ ; the coincidence is special to homogeneous powers and fails for $x^2 + 1$ .
Symmetry is invariance: on a domain symmetric about $0$ , $f$ is even exactly when reflecting the input fixes the graph ( $f(-x) = f(x)$ ). On that same domain, $f$ is odd exactly when reflecting the input matches reflecting the output ( $f(-x) = -f(x)$ ).

Transformations of Graphs

Learning goals

Two places a number can act

Why the inside runs backward

The master formula

Order matters, and the two sides disagree

When the order of two moves changes the result

One graph, many transformations

Symmetry is a transformation that does nothing

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)