Transformations — vertical - horizontal shifts, reflections, stretches - compressions: Question bank

Before we start, one reminder in plain words: the input of a function is the number you feed in (the $x$ ), and the output is the number that comes out (the height, the $y$ ). Every trap below is really about one question — did we touch the input or the output?

Definition The four parameters — what each letter means

Every trap on this page lives inside one master formula: $g (x) = a f (b (x - h)) + k$ Read it strictly from the input side outward. Here is what each letter is and where it acts:

$h$ = horizontal shift. It sits next to $x$ inside the function, so it moves the graph left/right. Positive $h$ → right, because $x - h$ needs a bigger $x$ to undo the subtraction.
$b$ = horizontal scale. It multiplies $x$ inside, so it squeezes/stretches sideways. $∣ b ∣ > 1$ compresses, $0 < ∣ b ∣ < 1$ stretches, and a negative $b$ also mirrors across the $y$ -axis.
$a$ = vertical scale. It multiplies the output $f (\dots)$ , so it stretches/compresses up-and-down. $∣ a ∣ > 1$ stretches, $0 < ∣ a ∣ < 1$ compresses, and a negative $a$ also mirrors across the $x$ -axis.
$k$ = vertical shift. It is added to the output, so it lifts/drops the graph. Positive $k$ → up.

Rule of thumb: letters touching $x$ (that's $h$ and $b$ ) are horizontal and behave in reverse; letters touching the output (that's $a$ and $k$ ) are vertical and behave as written. The figure below shows all four acting on a single parabola.

Figure — Transformations — vertical - horizontal shifts, reflections, stretches - compressions

True or false — justify

Each of these looks obviously true or obviously false. The catch is in the reason, not the verdict.

f (x) + 3

shifts the graph up 3, and

f (x + 3)

shifts it right 3.

False on the second part —

f (x + 3)

shifts left 3. Adding inside the bracket makes the function reach its old behaviour at a smaller

x

, so the graph slides toward the negatives (left).

For

g (x) = f (2 x)

, a bigger multiplier means a wider, more stretched-out graph.

False. Multiplying the input by

2

makes the function race through its values in half the horizontal distance, so it compresses toward the

y

-axis — bigger inside multiplier means narrower.

g (x) = - f (x)

and

g (x) = f (- x)

are the same transformation.

False.

- f (x)

negates the output (flip over the

x

-axis, up↔down);

f (- x)

negates the input (flip over the

y

-axis, left↔right). Same minus sign, opposite mirror.

Multiplying a function by

a = - 2

only reflects it, nothing else.

False.

a = - 2

does two things at once: reflect across the

x

-axis (from the minus) and vertically stretch by factor

2

(from the size). Sign and magnitude act independently.

The transformation

f (x) + k

never changes the domain of the function.

True. Adding

k

changes only outputs (heights), so the set of allowed inputs is untouched. Only horizontal transformations can move the domain — see Domain and Range.

A vertical stretch by factor

3

moves the

x

-intercepts.

False. At an

x

-intercept the output is

0

, and

3 \times 0 = 0

, so those points are pinned in place. Only points off the

x

-axis get pulled away.

For

g (x) = f (x - h)

with

h > 0

, every point

(x, y)

becomes

(x + h, y)

True. To reproduce the old output you now need a larger input, so each point slides right by

h

. The "

x - h

" in the formula and "

x + h

" for the point moving are two sides of the same compensation.

Reflecting

x

across the

y

-axis gives a graph with the same domain.

False.

x

lives on

[0, \infty)

;

- x

lives on

(- \infty, 0]

. A

y

-axis reflection flips the domain across zero too — inputs move, so their allowed set moves.

g (x) = a f (b (x - h)) + k

can always be built regardless of the order you apply the four steps.

False. Because inputs are processed first (composition works inside-out), horizontal steps and vertical steps each have their own required order. Do horizontals before verticals, and inside the bracket handle the shift before the scale — see Function Composition. The worked ordering in the "Why questions" section below shows exactly what goes wrong if you don't.

Spot the error

A worked solution is shown with one flaw. Find it and say why it breaks.

"To graph

y = (2 x - 6)^{2}

, I read off

b = 2

and

h = - 6

, so I compress by 2 and shift left 6."

The input isn't in the form

b (x - h)

yet. Factor first:

2 x - 6 = 2 (x - 3)

, so

h = + 3

(shift right 3), not

- 6

. Never read

h

off before factoring the coefficient out.

g (x) = f (3 x)

stretches the graph horizontally because the number

3

is large."

Backwards. Inside multipliers invert their effect: input

\times 3

means the graph hits each output three times sooner, so it compresses. Large inside number → narrow graph.

"For

cos (2 x)

the amplitude changes from 1 to 2."

The

2

is inside the cosine, so it changes the period (halves it), not the amplitude. Amplitude only changes when a number multiplies the output, like

2 cos (x)

f (x + 2) - 5

shifts right 2 and down 5."

Down 5 is right, but

x + 2

shifts left 2, not right. The outside subtraction behaves as expected; the inside addition flips direction.

"To turn

sin (x)

into

sin (\frac{x}{2})

I compress it, doubling the frequency."

A factor of

\frac{1}{2}

inside means

b = \frac{1}{2} < 1

, which stretches horizontally and halves the frequency (period doubles to

4 π

). Small inside multiplier → wider graph.

"Since

g (x) = - 3 f (x) + 1

reflects and shifts, I'll add 1 first, then multiply by

- 3

Wrong order. The

+ 1

is the outermost operation, so it happens last. Applying it first would give

- 3 (f (x) + 1)

, a different function. Follow the composition from inside out.

∣ f (x) ∣

is just a vertical reflection of

f (x)

Only the parts below the

x

-axis get reflected upward; parts already above stay put. It's a piecewise fold, not a full flip — see Absolute Value Functions and Piecewise Functions.

Why questions

The "obvious" answer is often the wrong instinct. Explain the real mechanism.

Why does subtracting inside the bracket,

f (x - h)

, move the graph right instead of left?

Because the graph plots outputs against inputs: to recreate the output that used to appear at input

a

, you must now feed in

a + h

(so that

a + h - h = a

). Larger required input = rightward slide.

Why do vertical transformations feel "intuitive" while horizontal ones feel reversed?

Vertical transformations act on the output after the function runs, so "

+ k

" literally adds height — no inversion. Horizontal ones act on the input before the function runs, so you must compensate for what the function will do, which flips the direction.

Why must horizontals be done before verticals — walk me through what breaks otherwise?

Take

f (x) = x^{2}

and build

g (x) = 2 f (x - 1) = 2 (x - 1)^{2}

. Correct order: shift right 1 first, sending

(0, 0) \to (1, 0)

, then stretch vertically by 2, keeping it at

(1, 0)

— the vertex lands at

x = 1

. If instead you stretch first to get

2 x^{2}

and then try to "shift right 1" by replacing

x

with

x - 1

, you get

2 (x - 1)^{2}

— which happens to agree here, but the shift you actually applied moved the input, not the already-scaled output. The safe rule is: inside-the-bracket steps (

h

then

b

) act on

x

and must be settled before the outside steps (

a

then

k

) act on the finished output. See Function Composition.

Why, inside the bracket, must the shift

h

be handled conceptually together with (not after) the scale

b

Because

b (x - h)

scales the already-shifted coordinate. Compare

f (2 (x - 1))

(shift right 1, then compress) with

f (2 x - 1) = f (2 (x - \frac{1}{2}))

(compress, effectively shifting right only

\frac{1}{2}

). Same-looking numbers, different graphs — which is exactly why you factor to read the true

h

from

b (x - h)

Why is

a < 0

described as "reflect and stretch" rather than one single move?

A negative

a

carries two pieces of information: its sign (a mirror across the

x

-axis) and its size

∣ a ∣

(the scaling factor). These are logically separate, so we name both. For

∣ a ∣ = 1

it's a pure reflection.

Why does a vertical stretch leave the period of

sin (x)

unchanged?

Period is a horizontal property (how far along

x

before it repeats), but a vertical stretch only touches outputs. Nothing about the input's spacing changes, so the wave repeats at the same rate — just taller. See Trigonometric Functions.

Why does reflecting a function's input also reflect its inverse relationship in a specific way?

Because swapping input for output is what an inverse does, so input-side transformations of

f

correspond to output-side transformations of

f^{- 1}

. The "inside vs outside" roles trade places — explored further in Inverse Functions.

Why can't we always write "

f (b x - c)

" without factoring before identifying the shift?

Take

f (3 x - 6)

. Read naively, you might call the shift

6

. But factor:

3 x - 6 = 3 (x - 2)

, so the true shift is

h = 2

, not

6

— the

b = 3

was hiding inside. In general

b x - c = b (x - \frac{c}{b})

, so the real shift is

\frac{c}{b}

; skipping the factor step misidentifies both the distance and (if

b < 0

) the direction.

Edge cases

Every quadrant, every sign, every degenerate value. These are the boundaries the rules quietly depend on.

What does

g (x) = a f (x)

do when

a = 0

It collapses the entire graph onto the line

y = 0

(the

x

-axis), because every output becomes

0 \cdot f (x) = 0

. This is why the definition requires

a \neq = 0

— at

a = 0

the function is destroyed, not transformed.

What does

g (x) = f (b x)

do when

b = 0

The input becomes a constant

0

for every

x

, so

g (x) = f (0)

is a flat horizontal line. All information about

x

is lost, which is why

b \neq = 0

is required.

For a general negative

b

, say

g (x) = f (- 2 x)

, what two things happen at once and in which order do they matter?

The magnitude

∣ b ∣ = 2

compresses horizontally (points move from

a

a /2

), and the minus sign reflects across the

y

-axis (points then flip to

- a /2

). Combined, a point at input

a

lands at

- a /2

: for

b < 0

every point

(x, y)

becomes

(x / b, y)

, which folds the graph to the opposite side and squeezes/stretches it by

1/∣ b ∣

— the special case

b = - 1

is just the pure reflection with no scaling.

What does

g (x) = f (b x)

do when

- 1 < b < 0

, e.g.

b = - \frac{1}{2}

Two effects: reflect across the

y

-axis (from the minus) and stretch horizontally by

2

(since

∣ b ∣ < 1

). A point at

a

maps to

a / (- \frac{1}{2}) = - 2 a

, farther out and flipped — showing negative

b

with small magnitude still reflects but widens rather than narrows.

What happens to

y = x^{2}

under

g (x) = f (- x)

, and why does it look like nothing changed?

Because

x^{2}

is symmetric about the

y

-axis (even function), reflecting across that axis maps it onto itself. The transformation happened; the graph just happens to be its own mirror image.

What is the effect of

g (x) = f (x - 0) + 0

Nothing — it's the identity transformation. Zero shift in both directions leaves every point

(x, y)

exactly where it was, a useful sanity check that the point-mapping rules reduce correctly.

∣ a ∣ = 1

but

a < 0

, is that a stretch, a compression, or neither?

Neither in size — it's a pure reflection across the

x

-axis, since multiplying by

- 1

keeps each output's distance from the axis but flips its sign. Only

∣ a ∣ \neq = 1

adds actual stretching or compressing.

For

g (x) = f (b x)

with

0 < b < 1

, why does the domain grow rather than shrink?

A point at input

a

moves to

a / b

, and dividing by a number below

1

makes it larger in magnitude, spreading points outward. The stretch pulls the domain wider — matching that small

b

means horizontal stretch.

Which parameter choices in

g (x) = a f (b (x - h)) + k

leave every point of

y = sin (x)

fixed, and how do we check?

Only

a = 1, b = 1, h = 0, k = 0

(the identity). Systematic check by testing the four transformations one at a time:

k \neq = 0

moves the point

(0, 0)

off the axis;

h \neq = 0

moves the zero at

x = 0

x = h

;

a \neq = 1

changes the peak height from

1

;

b \neq = 1

changes the period from

2 π

. Since each nonzero/non-unit parameter visibly relocates at least one identified point, no non-trivial choice can fix the whole curve.