Why does f(x − 2) shift the graph to the right rather than the left?

Because it is the argument that changes, not the value. The new function does at x what the old one did at x − 2, so to recover the old value you have to travel two units further right. A quick check: the old function had its vertex at zero, and the new one has it where x − 2 = 0, that is at x = 2.

What is the difference between −f(x) and f(−x)?

The first flips the sign of the value, so it reflects the graph in the x-axis (what was high is now low). The second flips the sign of the argument, so it reflects the graph in the y-axis (what was on the right is now on the left). For the parabola y = x² the second does nothing at all, because the function is even, while the first turns it arms-down.

In what order do you apply a combined transformation?

Whatever acts on the argument first, whatever acts on the value second — the same order in which you would evaluate the function. In y = 2·f(x − 3) + 1 you shift 3 right, then stretch twice vertically, then lift by 1. Swapping the stretch and the vertical shift gives a different graph, because 2·(f(x) + 1) is not 2·f(x) + 1.

What does the coefficient a do in y = a·f(x)?

It stretches the graph along the y-axis: every point moves a times further from the x-axis. For |a| > 1 the graph gets taller, for 0 < |a| < 1 it flattens, and for a < 0 a reflection in the x-axis comes with it. Points on the x-axis stay put, because zero times anything is still zero.

How do you turn y = sin x into y = a·sin(bx + c)?

In three moves. The coefficient a sets the amplitude, stretching the graph vertically to height |a|. The coefficient b squeezes it horizontally: the period 2π becomes 2π/|b|. The constant c shifts the whole thing horizontally by −c/b, because bx + c = b·(x + c/b) — which is why the phase shift is c divided by b rather than c itself.

What is the period of y = 3·sin 4x?

The fundamental period is 2π/4 = π/2. The 3 stands in front of the function rather than beside the argument, so it does not touch the period — it only sets the amplitude to 3. The graph therefore completes a full cycle four times faster than a plain sine, reaching from −3 to 3.

Does shifting a graph change the domain?

A vertical shift does not, a horizontal one does. The function y = f(x) + q has the same domain as f, because the argument was untouched. The function y = f(x − p) has the domain moved by p: if f was defined for x > 0, the new one is defined for x > p.

How do you know the transformed curve is still the graph of a function?

Because each of these transformations moves points vertically or horizontally and never stacks two of them above one another. The vertical-line test that passed before the transformation passes after it too. The exception would be a reflection in the line y = x, which swaps the roles of the axes — and that is the inverse function, a topic of its own.

Intermediate level

Transforming a graph

Q: How do you turn y = sin x into y = a·sin(bx + c)?

In three moves. The coefficient a sets the amplitude, stretching the graph vertically to height |a|. The coefficient b squeezes it horizontally: the period 2π becomes 2π/|b|. The constant c shifts the whole thing horizontally by −c/b, because bx + c = b·(x + c/b) — which is why the phase shift is c divided by b rather than c itself.

Q: What is the period of y = 3·sin 4x?

The fundamental period is 2π/4 = π/2. The 3 stands in front of the function rather than beside the argument, so it does not touch the period — it only sets the amplitude to 3. The graph therefore completes a full cycle four times faster than a plain sine, reaching from −3 to 3.

Q: Does shifting a graph change the domain?

A vertical shift does not, a horizontal one does. The function y = f(x) + q has the same domain as f, because the argument was untouched. The function y = f(x − p) has the domain moved by p: if f was defined for x > 0, the new one is defined for x > p.

One curve, four moves: up, sideways, mirrored in an axis and stretched. See how a change in the formula turns into a movement of the curve, why f(x − 2) shifts the graph to the right rather than the left, and how a plain sine becomes y = a·sin(bx + c) with any amplitude and period you like.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Quadratic functionA quadratic function f(x) = ax² + bx + c draws a parabola. Meet the vertex and the axis of symmetry, the three forms of the formula — general, canonical and factored — and how the sign of a decides between a minimum and a maximum.
Trigonometric graphsA sine wave is the unit circle unrolled along an axis. See where the period of 2π comes from, why the cosine is a shifted sine, and what happens to the graph of the tangent where the function does not exist.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Tuning an instrument
A tuning fork sounds A4 at 440 Hz, that is the wave y = a·sin(2π·440·t). The same string pulled tighter sounds 466 Hz — the coefficient beside t grows, so the period drops from 1/440 ≈ 2.27 ms to 1/466 ≈ 2.15 ms. The tuner changes no shape, only one coefficient in the formula; plucking harder changes the amplitude alone, which is the loudness.
Tides in a harbour
The water level in Gdańsk is modelled by h(t) = 0.3·sin(0.5t) + 2.4, with t in hours. The amplitude 0.3 m is half the difference between high and low water, and the constant 2.4 m is the upward shift, that is the mean sea level. A captain substitutes t = 3 and gets h = 0.3·sin 1.5 + 2.4 ≈ 2.70 m — the depth under the keel three hours from now.
A ball kicked on a pitch
A ball kicked off the ground follows the parabola y = −0.05x² + x. The same kick taken from a 2 m platform is the same curve moved up: y = −0.05x² + x + 2. The vertex rises from 5 m to 7 m and the range grows from 20 m to about 21.8 m — the coach gets the answer without recomputing the trajectory.
Seasonal sales analysis
Ice-cream sales across a year are modelled by S(m) = 120·sin(0.52·(m − 4)) + 200 thousand zloty, where m is the month number. The shift by 4 puts the peak in July; the amplitude 120 states the swing about the mean of 200 thousand. An analyst substitutes m = 1 and forecasts S ≈ 200 − 120·0.84 ≈ 99 thousand for January, less than half the mean.

All formulas

Shift along the y-axis
$y = f(x) + q$
up for q > 0, down for q < 0
Shift along the x-axis
$y = f(x - p)$
to the right for p > 0 — against the sign in the formula
Translation by a vector
$y = f(x - p) + q$
\vec{u} = [p,\, q]
Reflection in the x-axis
$y = -f(x)$
flips the sign of the value
Reflection in the y-axis
$y = f(-x)$
flips the sign of the argument
Vertical stretch
$y = a \cdot f(x)$
a times further from the x-axis
Horizontal squeeze
$y = f(b x)$
b times closer to the y-axis
The general sinusoid
$y = a \sin(bx + c)$
amplitude |a|, period \tfrac{2\pi}{|b|}, shifted by -\tfrac{c}{b}

Not every curve has to be drawn from scratch. Most graphs you will meet are a handful of familiar shapes shifted, mirrored or stretched — and every one of those moves is visible in the formula.

Shifting along the y-axis

The simplest case: add a number to a finished formula.

y = f(x) + q

Every value grows by $q$ , so the whole graph rises by $q$ — up for $q > 0$ , down for $q < 0$ . The shape does not change, because the same amount was added to every value.

For the quadratic function $f(x) = x^2$ , the formula $y = x^2 + 3$ gives the same parabola with its vertex at $(0,\, 3)$ instead of at the origin.

Shifting along the x-axis

Here the argument changes, and this is where mistakes happen:

y = f(x - p)

The graph moves $p$ units to the right — despite the minus sign. The reason is simple once you have seen it: the new function does at $x$ exactly what the old one did at $x - p$ . To recover the old value at zero, you have to travel all the way to $x = p$ .

formula	what happens
$y = f(x - 3)$	three right
$y = f(x + 3)$	three left
$y = f(x) + 3$	three up
$y = f(x) - 3$	three down

The two middle rows look alike and do opposite things. The sign beside the argument works against intuition; the sign beside the value works with it.

Translation by a vector

Both moves at once make a translation by the vector $\vec{u} = [p,\, q]$ :

y = f(x - p) + q

The parabola y = x² and its image under the translation by [2, 3]. The arrow runs from the old vertex to the new one — y = (x − 2)² + 3 says exactly the same thing in letters.

Multiplied out, $(x-2)^2 + 3 = x^2 - 4x + 7$ — the same function, but the general form hides the translation. The vertex form is precisely the spelling in which the shift is written out.

The graph of f(x) = x² is translated by the vector [−3, 2]. Give the formula of g.

Reflections

Two transformations, two different minus signs:

y = -f(x) \qquad \text{and} \qquad y = f(-x)

The first flips the sign of the value — the point $(x,\, y)$ goes to $(x,\, -y)$ , so the graph is reflected in the x-axis. The second flips the sign of the argument — the point $(x,\, y)$ goes to $(-x,\, y)$ , a reflection in the y-axis.

function	$-f(x)$	$f(-x)$
$x^2$	$-x^2$ (arms down)	$x^2$ (unchanged — an even function)
$2x + 1$	$-2x - 1$	$-2x + 1$
$\sqrt{x}$	$-\sqrt{x}$	$\sqrt{-x}$ (domain $x \le 0$ )

The middle row shows why it pays to compute rather than guess: for $f(x) = 2x+1$ the two reflections give different lines.

Given f(x) = x² − 4x, find the formula of g(x) = f(−x).

Stretching and squeezing

Instead of adding, multiply — and again the result depends on which side the multiplier stands.

y = a \cdot f(x) \qquad \text{and} \qquad y = f(b x)

$y = a \cdot f(x)$ stretches the graph along the y-axis: every point moves $a$ times further from the x-axis. The zeros stay where they were, because $a \cdot 0 = 0$ .
$y = f(bx)$ squeezes the graph along the x-axis: what the function used to do over a unit interval it now fits into an interval of length $\tfrac{1}{b}$ . The intercept on the y-axis stays put.

Once more: the multiplier on the value behaves as you expect, the multiplier on the argument does the opposite. A 2 in front of the function makes the graph twice as tall; a 2 beside $x$ makes it twice as narrow.

The general sinusoid

All four moves at once show up best on the sine graph, because a plain $\sin x$ has such a recognisable shape:

y = a \sin(bx + c) + d

coefficient	what it sets	value
$a$	amplitude	$\lvert a \rvert$
$b$	period	$\dfrac{2\pi}{\lvert b \rvert}$
$c$	phase shift	$-\dfrac{c}{b}$
$d$	midline	$y = d$

A plain sine and the sinusoid y = 2·sin 2x. The crests reach 2 instead of 1 (the amplitude), and a full cycle fits into π instead of 2π (the period).

The phase shift is $c$ divided by $b$ , not $c$ itself — only after factoring $b$ out of the bracket can you see how far the argument really travels:

\sin(2x - \pi) = \sin\!\left(2\left(x - \tfrac{\pi}{2}\right)\right)

The graph moves $\tfrac{\pi}{2}$ to the right, not $\pi$ .

Give the amplitude, the period and the phase shift of y = 3 sin(2x + π).

The same three coefficients describe every wave: sound, mains voltage and the water level in a harbour differ only in their values.

Practice

The set checks four things: the formula after a shift or a reflection, the image of a point of the graph (compute $f$ first, then move the point), the period of a sinusoid and its value at a point. Write the formula in whichever spelling suits you — (x−2)^2+3 and x^2−4x+7 are both accepted, because they are the same function. Describing a transformation in words ("a translation by the vector…") stays with the worked examples above: its answer is a sentence, not a number.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

f(x) = x²; g(x) = f(x + 1) + 4; g(x) =

Common mistakes

Shifting left for $f(x - p)$ — a minus beside the argument means a move to the right; check it on the vertex or on a zero.
Confusing $-f(x)$ with $f(-x)$ — the first reflects in the x-axis, the second in the y-axis; for an odd function they agree, which is exactly what lulls you into not checking.
Reading the phase shift off $c$ — only $-\tfrac{c}{b}$ says how far the graph travels.
Believing $a$ in $y = a\sin bx$ changes the period — it changes the amplitude; the period depends on $b$ alone.
Swapping a stretch and a vertical shift — $2 \cdot f(x) + 1$ and $2 \cdot (f(x) + 1)$ are two different graphs.
Forgetting the domain after a reflection — $\sqrt{-x}$ is defined for $x \le 0$ , however innocent the formula looks.

Formula card

Topic: Transforming a graph

Shift along the y-axis
$y = f(x) + q$
up for q > 0, down for q < 0
Shift along the x-axis
$y = f(x - p)$
to the right for p > 0 — against the sign in the formula
Translation by a vector
$y = f(x - p) + q$
\vec{u} = [p,\, q]
Reflection in the x-axis
$y = -f(x)$
flips the sign of the value
Reflection in the y-axis
$y = f(-x)$
flips the sign of the argument
Vertical stretch
$y = a \cdot f(x)$
a times further from the x-axis
Horizontal squeeze
$y = f(b x)$
b times closer to the y-axis
The general sinusoid
$y = a \sin(bx + c)$
amplitude |a|, period \tfrac{2\pi}{|b|}, shifted by -\tfrac{c}{b}

The parabola y = x² and the same parabola translated by the vector [2, 3]. The arrow shows where the vertex went: y = (x − 2)² + 3 reads as "two right, three up".

A sine and the sinusoid y = 2·sin(2x). The coefficient in front of the function lifts the crests to 2; the coefficient beside x cuts the period from 2π to π — two coefficients, two different effects.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min

Transforming a graph

Before you start

Where this is used

All formulas

Shifting along the y-axis

Shifting along the x-axis

Translation by a vector

Reflections

Stretching and squeezing

The general sinusoid

Practice

Practice

Common mistakes

Formula card

Frequently asked questions

Why does f(x − 2) shift the graph to the right rather than the left?

What is the difference between −f(x) and f(−x)?

In what order do you apply a combined transformation?

What does the coefficient a do in y = a·f(x)?

How do you turn y = sin x into y = a·sin(bx + c)?

What is the period of y = 3·sin 4x?

Does shifting a graph change the domain?

How do you know the transformed curve is still the graph of a function?

Related articles

Why a function's graph can never double back