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

1. Vertical Transformations (Outside the function)

Vertical Shifts

Derivation from first principles: Start with any point $P = (a, f(a))$ on the original graph. After transformation, the new function gives us: $g(a) = f(a) + k$

The new point is $P' = (a, f(a) + k)$ . The $x$ -coordinate stays the same, but the $y$ -coordinate increases by $k$ . This is literally adding $k$ to the height of every point.

WHY does this work? Because we're modifying the OUTPUT. Whatever $f(x)$ spits out, we add $k$ to it. Since output = height on a graph, we're adding to the height.

Vertical Stretch/Compression

Derivation: For point $(c, f(c))$ , the new function gives: $g(c) = a \cdot f(c)$

So $y$ -coordinate is multiplied by $a$ . The $x$ -coordinate is unchanged.

WHY does $|a| > 1$ stretch? Because we're multiplying the output by a number bigger than 1, making the output larger. If $f(x) = 2$ , then $3f(x) = 6$ — that's farther from the $x$ -axis. If $f(x) = -2$ , then $3f(x) = -6$ — also farther from the axis.

Reflection Across $x$ -axis

WHY? We're negating the output. If the graph was above the $x$ -axis, it flips below. The $x$ -intercepts stay fixed (because $-0 = 0$ ).

2. Horizontal Transformations (Inside the function)

Horizontal Shifts

Derivation from "compensation" logic: The equation $y = f(x - h)$ can be rewritten by substituting $u = x - h$ , so $x = u + h$ .

When $u = 0$ (the original position where something happened), we now need $x = h$ for the same thing to happen. So the entire graph shifts RIGHT by $h$ .

Formal proof: If $(a, f(a))$ is on the original graph, then for the transformed graph: $g(x) = f(x - h)$ We want $g(x) = f(a)$ , so: $f(x - h) = f(a) \implies x - h = a \implies x = a + h$

The point is now at $(a + h, f(a))$ — shifted right by $h$ .

Horizontal Stretch/Compression

Derivation: If $(a, f(a))$ is on the original graph, for the transformed graph: $g(x) = f(bx)$ We want $g(x) = f(a)$ : $f(bx) = f(a) \implies bx = a \implies x = \frac{a}{b}$

The point is now at $(\frac{a}{b}, f(a))$ .

WHY compression with $b > 1$ ? Because $\frac{a}{b}$ is smaller than $a$ when $b > 1$ . The function "speeds up" — it reaches the same outputs faster (in less horizontal distance).

Reflection Across $y$ -axis

WHY? We're negating the input. What used to happen at $x = 3$ now happens at $x = -3$ (because we need $-x = 3$ , so $x = -3$ ).

3. Combined Transformations

WHY this order? Because of function composition. Work from the inside out. The input $x$ first gets shifted, then scaled, then the output gets scaled, then shifted.

Worked example Example 5: Complex transformation

Transform $f(x) = \sqrt{x}$ into $g(x) = -2\sqrt{3(x + 1)} + 4$ .

Step-by-step:

Start with $\sqrt{x}$ (domain $[0, \infty)$ , starting point $(0, 0)$ )
$\sqrt{x + 1}$ : shift LEFT1 unit → starting point $(-1, 0)$
$\sqrt{3(x + 1)}$ : compress horizontally by factor $\frac{1}{3}$ (graph runs3× faster)
$-2\sqrt{3(x + 1)}$ : reflect across $x$ -axis and stretch vertically by 2
$-2\sqrt{3(x + 1)} + 4$ : shift UP 4 units

Why each step?

Step 2: $x + 1 = 0 \implies x = -1$ for the starting point
Step 3: Point that was at $x = 3$ (since $\sqrt{3} \approx 1.73$ ) now at $x = 0$ (because $3(0 + 1) = 3$ )
Step 4: $y$ -values flip and double in magnitude; $(0, \sqrt{3})$ becomes $(0, -2\sqrt{3})$
Step 5: Add 4 to all $y$ -values; starting point goes from $(-1, 0)$ to $(-1, 4)$

Final key points: Start at $(-1, 4)$ , passes through $(0, 4- 2\sqrt{3}) \approx (0, 0.54)$ , decreasing due to reflection.

Common Mistakes

Active Recall

Recall Explain to a 12-year-old

Imagine you have a picture (that's your function graph). Now you want to move it around or change its size, but WITHOUT drawing a new picture from scratch.

Moving up/down is easy: just add or subtract a number at the end. Add 5? Every point goes up 5 steps.

Moving left/right is TRICKY. It's backwards! If you see "minus 3" INSIDE (like $f(x-3)$ ), the picture actually moves RIGHT by 3. Why? Think of it like this: "I'm asking the function to do its thing3 steps earlier." If something used to happen at step 5, you now need step 8 to see it (because 8 - 3 = 5).

Making it taller/shorter: multiply the whole function by a number. Times 2? Everything gets twice as tall. Times 0.5? Everything gets squished to half height.

Making it wider/narrower: Put a number in front of $x$ INSIDE the function. This is also backwards! Times 2 INSIDE (like $f(2x)$ ) makes it NARROWER, not wider, because the function "speeds up" — it does its thing twice as fast, so it finishes in half the distance.

Flipping: Negative sign OUTSIDE flips upside-down. Negative sign INSIDE (on the $x$ ) flips left-to-right.

Connections

Parent Functions — these are what we transform
Function Composition — transformations are composed functions
Domain and Range — transformations change these
Trigonometric Functions — period changes use horizontal compression
Inverse Functions — reflection across $y = x$ relates to these concepts
Absolute Value Functions — transformations create all variations
Piecewise Functions — each piece transforms independently

#flashcards/maths

What happens to the graph of $f(x)$ when you compute $f(x) + k$ for $k > 0$ ? :: The graph shifts vertically UPWARD by $k$ units. Every point $(x, y)$ becomes $(x, y+k)$ .

What happens to the graph of

f(x)

when you compute

f(x - h)

for

h > 0

The graph shifts horizontally to the RIGHT by

h

units. Every point

(x, y)

becomes

(x+h, y)

. (Counterintuitive: minus means right!)

For

g(x) = af)

, when does vertical stretch occur?

When

|a| > 1

. The graph is pulled away from the

x

-axis. Every

y

-coordinate is multiplied by

a

For

g(x) = f(bx)

, when does horizontal compression occur?

When

|b| > 1

. The graph is squeezed toward the

y

-axis. Every

x

-coordinate is divided by

b

(factor of

\frac{1}{b}

What transformation does

g(x) = -f(x)

represent?

Reflection across the

x

-axis. Every point

(x, y)

becomes

(x, -y)

What transformation does $g(x) = f(-x)$ represent? :: Reflection across the $y$ -axis. Every point $(x, y)$ becomes $(-x, y)$ .

g(x) = af(b(x-h)) + k

, what is the correct order to apply transformations when sketching?

1) Horizontal shift by

h

, 2) Horizontal scale by

\frac{1}{b}

, 3) Vertical scale by

a

, 4) Vertical shift by

k

. (Inside-out order)

f(x) = x^2

and

g(x) = (x+3)^2

, which direction does the graph shift?

LEFT by 3 units. The

(x+3)

means subtract

-3

, so

h = -3

, giving a leftward shift.

For the function

h(x) = 3f(2x)

, what two transformations occur?

Horizontal compression by factor

\frac{1}{2}

(graph squezed) AND vertical stretch by factor 3 (graph pulled away from

x

-axis).

\sin(x)

has period

2\pi

, what is the period of

\sin(3x)

\frac{2\pi}{3}

. The period formula is

\frac{\text{original period}}{|b|}

where

b

is the coefficient of

x

How do you find the new vertex of

f(x) = (x-h)^2 + k

from the parent

x^2

The vertex shifts from

(0, 0)

(h, k)

. Horizontal shift right by

h

, vertical shift up by

k

What changes when you apply

g(x) = 0.5f(x)

to a function?

Vertical compression by factor 0.5 (toward the

x

-axis). Every

y

-coordinate is halved. Range shrinks by factor 0.5.

For

g(x) = f(\frac{x}{3})

, what happens to the graph?

Horizontal stretch by factor 3 (away from

y

-axis). The graph becomes wider. If a feature was at

x=1

, it's now at

x=3

Why does

f(x-2)

shift the graph RIGHT instead of left?

Because of compensation: to get the output that used to occur at

x=0

, you now need

x=2

(since

2-2=0

). The entire graph moves right.

If the domain of

f(x)

[0, 5]

, what is the domain of

f(x-3)

[3, 8]

. Every domain value shifts right by 3 units. The left endpoint

0 \to 3

, right endpoint

5 \to 8

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Chalo isko simple tarike se samajhte hain. Kisi bhi function $f(x)$ ko ek machine ki tarah socho — input daalte ho, output nikalta hai. Transformations bas is machine ke saath do jagah chhed-chhaad karte hain: ya toh output pe (jo bahar aata hai), ya input pe (jo andar jaata hai). Jab tum output modify karte ho jaise $f(x)+k$ ya $af(x)$ , toh graph mein vertical (upar-neeche) change hota hai — aur ye seedha-saada hota hai, jaise expect karte ho waisa hi. Point $(x,y)$ ban jaata hai $(x, y+k)$ , matlab har point ki height badal jaati hai.

Ab asli maza ya twist input side pe aata hai. Jab tum input ke saath khelte ho jaise $f(x+h)$ ya $f(bx)$ , toh change horizontal hota hai, lekin ulti direction mein — isko hi "inside-out paradox" kehte hain. Yani $f(x+2)$ graph ko right nahi, balki left shift karta hai. Beginners yahin confuse hote hain, isliye is baat ko dil se yaad rakhna. Aur agar aage minus sign lag jaaye jaise $-f(x)$ , toh graph $x$ -axis ke across flip ho jaata hai, jabki $|a|>1$ hone pe stretch aur $0<|a|<1$ hone pe compression hota hai.

Ye cheez itni important kyun hai? Kyunki exam mein jitne bhi complicated graphs aate hain, woh sab basically ek simple parent function (jaise $x^2$ ya $\sin x$ ) hote hain jo bas shift, stretch, flip ya compress kiye gaye hote hain. Agar tumne ye chaar transformations acche se pakad liye, toh tumhe sau-sau shapes ratne ki zaroorat hi nahi — koi bhi function dekh ke turant graph sketch kar sakte ho. Yeh ek chhota sa concept hai par pura Functions chapter aur aage Calculus mein bahut kaam aata hai.

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

1. Vertical Transformations (Outside the function)

Vertical Shifts

Vertical Stretch/Compression

Reflection Across $x$ -axis

2. Horizontal Transformations (Inside the function)

Horizontal Shifts

Horizontal Stretch/Compression

Reflection Across $y$ -axis

3. Combined Transformations

Common Mistakes

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Functions

Connections