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

2.2.7 · Maths › Functions

1. Vertical Transformations (Function ke bahar)

Vertical Shifts

First principles se derivation: Original graph par koi bhi point $P = (a, f (a))$ lo. Transformation ke baad, naya function deta hai: $g (a) = f (a) + k$

Naya point hai $P^{'} = (a, f (a) + k)$ . $x$ -coordinate same rehta hai, lekin $y$ -coordinate $k$ se badh jaata hai. Yeh literally har point ki height mein $k$ add karna hai.

WHY yeh kaam karta hai? Kyunki hum OUTPUT modify kar rahe hain. Jो bhi $f (x)$ output deta hai, hum usme $k$ add karte hain. Kyunki output = graph par height, hum height mein add kar rahe hain.

Vertical Stretch/Compression

Derivation: Point $(c, f (c))$ ke liye, naya function deta hai: $g (c) = a \cdot f (c)$

To $y$ -coordinate $a$ se multiply ho jaata hai. $x$ -coordinate unchanged rehta hai.

WHY $∣ a ∣ > 1$ stretch karta hai? Kyunki hum output ko 1 se bade number se multiply kar rahe hain, output ko bada bana rahe hain. Agar $f (x) = 2$ hai, toh $3 f (x) = 6$ — yeh $x$ -axis se zyada door hai. Agar $f (x) = - 2$ hai, toh $3 f (x) = - 6$ — yeh bhi axis se zyada door hai.

$x$ -axis ke across Reflection

WHY? Hum output ko negate kar rahe hain. Agar graph $x$ -axis ke upar tha, woh neeche flip ho jaata hai. $x$ -intercepts fixed rehte hain (kyunki $- 0 = 0$ ).

2. Horizontal Transformations (Function ke andar)

Horizontal Shifts

"Compensation" logic se derivation: Equation $y = f (x - h)$ ko $u = x - h$ substitute karke rewrite kiya ja sakta hai, toh $x = u + h$ .

Jab $u = 0$ (original position jahan kuch hua tha), ab wahi cheez hone ke liye humein $x = h$ chahiye. Toh poora graph RIGHT by $h$ shift ho jaata hai.

Formal proof: Agar $(a, f (a))$ original graph par hai, toh transformed graph ke liye: $g (x) = f (x - h)$ Hum chahte hain $g (x) = f (a)$ , toh: $f (x - h) = f (a) ⟹ x - h = a ⟹ x = a + h$

Point ab $(a + h, f (a))$ par hai — $h$ se right shift hua.

Horizontal Stretch/Compression

Derivation: Agar $(a, f (a))$ original graph par hai, transformed graph ke liye: $g (x) = f (b x)$ Hum chahte hain $g (x) = f (a)$ : $f (b x) = f (a) ⟹ b x = a ⟹ x = \frac{a}{b}$

Point ab $(\frac{a}{b}, f (a))$ par hai.

WHY $b > 1$ ke saath compression hoti hai? Kyunki $\frac{a}{b}$ , $a$ se chhota hota hai jab $b > 1$ ho. Function "speed up" karta hai — woh same outputs tak zyada jaldi pahuncha jaata hai (kam horizontal distance mein).

$y$ -axis ke across Reflection

WHY? Hum input ko negate kar rahe hain. Jo pehle $x = 3$ par hota tha ab $x = - 3$ par hota hai (kyunki humein $- x = 3$ chahiye, toh $x = - 3$ ).

3. Combined Transformations

WHY yeh order? Kyunki function composition ki wajah se. Andar se bahar kaam karo. Input $x$ pehle shift hota hai, phir scale hota hai, phir output scale hota hai, phir shift hota hai.

Worked example Example 5: Complex transformation

$f (x) = x$ ko $g (x) = - 2 3 (x + 1) + 4$ mein transform karo.

Step-by-step:

$x$ se shuru karo (domain $[0, \infty)$ , starting point $(0, 0)$ )
$x + 1$ : LEFT 1 unit shift → starting point $(- 1, 0)$
$3 (x + 1)$ : $\frac{1}{3}$ factor se horizontally compress karo (graph 3× tez chalta hai)
$- 2 3 (x + 1)$ : $x$ -axis ke across reflect karo aur vertically 2 se stretch karo
$- 2 3 (x + 1) + 4$ : 4 units UPAR shift karo

Har step kyun?

Step 2: $x + 1 = 0 ⟹ x = - 1$ starting point ke liye
Step 3: Jo point $x = 3$ par tha (kyunki $3 \approx 1.73$ ) ab $x = 0$ par hai (kyunki $3 (0 + 1) = 3$ )
Step 4: $y$ -values flip aur magnitude mein double ho jaate hain; $(0, 3)$ ban jaata hai $(0, - 23)$
Step 5: Sabhi $y$ -values mein 4 add karo; starting point $(- 1, 0)$ se $(- 1, 4)$ ho jaata hai

Final key points: $(- 1, 4)$ par start, $(0, 4 - 23) \approx (0, 0.54)$ se guzarta hai, reflection ki wajah se decreasing.

Common Mistakes

Active Recall

Recall Ek 12-saal ke bacche ko explain karo

Imagine karo tumhare paas ek picture hai (woh tumhara function graph hai). Ab tum use idhar-udhar move karna chahte ho ya uska size change karna chahte ho, lekin BINA scratch se naya picture banaye.

Upar/neeche move karna easy hai: bas end mein ek number add ya subtract karo. 5 add kiya? Har point 5 steps upar jaata hai.

Left/right move karna TRICKY hai. Yeh ulta hai! Agar tum ANDAR "minus 3" dekhte ho (jaise $f (x - 3)$ ), picture actually RIGHT by 3 move karti hai. Kyun? Aise socho: "Main function ko apna kaam 3 steps pehle karne ke liye bol raha hun." Agar kuch step 5 par hota tha, toh ab tumhe step 8 chahiye use dekhne ke liye (kyunki 8 - 3 = 5).

Use tall/short banana: poori function ko ek number se multiply karo. Times 2? Sab kuch do guna tall ho jaata hai. Times 0.5? Sab kuch half height par squish ho jaata hai.

Use wider/narrow banana: $x$ ke aage ANDAR function mein ek number dalo. Yeh bhi ulta hai! Times 2 ANDAR (jaise $f (2 x)$ ) use NARROWER banata hai, wider nahi, kyunki function "speed up" ho jaata hai — woh apna kaam do guna tez karta hai, toh half distance mein finish karta hai.

Flipping: BAHAR negative sign upar-neeche flip karta hai. ANDAR negative sign (on the $x$ ) left-to-right flip karta hai.

Connections

Parent Functions — yahi hain jo hum transform karte hain
Function Composition — transformations composed functions hain
Domain and Range — transformations inhe change karte hain
Trigonometric Functions — period changes horizontal compression use karte hain
Inverse Functions — $y = x$ ke across reflection in concepts se related hai
Absolute Value Functions — transformations saari variations create karte hain
Piecewise Functions — har piece independently transform hota hai

#flashcards/maths

$f (x)$ ka graph kya hota hai jab tum $f (x) + k$ compute karte ho $k > 0$ ke liye? :: Graph vertically UPAR $k$ units shift ho jaata hai. Har point $(x, y)$ ban jaata hai $(x, y + k)$ .

f (x)

ka graph kya hota hai jab tum

f (x - h)

compute karte ho

h > 0

ke liye?

Graph horizontally RIGHT by

h

units shift ho jaata hai. Har point

(x, y)

ban jaata hai

(x + h, y)

. (Counterintuitive: minus matlab right!)

g (x) = a f (x)

ke liye, vertical stretch kab hoti hai?

Jab

∣ a ∣ > 1

. Graph

x

-axis se door kheencha jaata hai. Har

y

-coordinate

a

se multiply hota hai.

g (x) = f (b x)

ke liye, horizontal compression kab hoti hai?

Jab

∣ b ∣ > 1

. Graph

y

-axis ki taraf squeeze hota hai. Har

x

-coordinate

b

se divide hota hai (

\frac{1}{b}

ka factor).

g (x) = - f (x)

kaunsi transformation represent karta hai?

x

-axis ke across reflection. Har point

(x, y)

ban jaata hai

(x, - y)

$g (x) = f (- x)$ kaunsi transformation represent karta hai? :: $y$ -axis ke across reflection. Har point $(x, y)$ ban jaata hai $(- x, y)$ .

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

mein, sketching karte waqt transformations apply karne ka sahi order kya hai?

h

se horizontal shift, 2)

\frac{1}{b}

se horizontal scale, 3)

a

se vertical scale, 4)

k

se vertical shift. (Inside-out order)

Agar

f (x) = x^{2}

aur

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

, graph kis direction mein shift hota hai?

LEFT by 3 units.

(x + 3)

matlab

- 3

subtract karo, toh

h = - 3

, leftward shift deta hai.

Function

h (x) = 3 f (2 x)

ke liye, kaunsi do transformations hoti hain?

Horizontal compression by factor

\frac{1}{2}

(graph squeeze hota hai) AUR vertical stretch by factor 3 (graph

x

-axis se door kheencha jaata hai).

Agar

sin (x)

ka period

2 π

hai, toh

sin (3 x)

ka period kya hai?

\frac{2 π}{3}

. Period formula hai

\frac{original period}{∣ b ∣}

jahan

b

x

ka coefficient hai.

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

ka naya vertex parent

x^{2}

se kaise nikalte hain?

Vertex

(0, 0)

(h, k)

shift ho jaata hai.

h

se horizontally right shift,

k

se vertically upar shift.

Jab tum

g (x) = 0.5 f (x)

kisi function par apply karte ho toh kya change hota hai?

x

-axis ki taraf vertical compression by factor 0.5. Har

y

-coordinate half ho jaata hai. Range 0.5 factor se shrink ho jaati hai.

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

ke liye, graph ka kya hota hai?

Factor 3 se horizontal stretch (

y

-axis se door). Graph wider ho jaata hai. Agar koi feature

x = 1

par tha, ab

x = 3

par hoga.

f (x - 2)

graph ko LEFT ki jagah RIGHT kyun shift karta hai?

Compensation ki wajah se: woh output paane ke liye jo pehle

x = 0

par aata tha, ab tumhe

x = 2

chahiye (kyunki

2 - 2 = 0

). Poora graph right move karta hai.

Agar

f (x)

ka domain

[0, 5]

hai, toh

f (x - 3)

ka domain kya hai?

[3, 8]

. Har domain value 3 units right shift hoti hai. Left endpoint

0 \to 3

, right endpoint

5 \to 8