Persamaan garis y = x^2 + 3 didilatasi oleh [P(1,2), D(1/2)]. bayangan kurva adalah....

jawab

y = x² + 3

dilatasi [ P(a,b), D (k) ] = {P(1,2 ) , k = 1/2}

x' = k (x - a) + a
x' = 1/2 (x - 1) + 1
x' = 1/2 x  - 1/2  + 1
x' = 1/2 x  + 1/2  ..kali 2
2x' = x + 1
x  = 1 - 2x'.....(i)

y' = k(y-b) + b
y' = 1/2 (y - 2) + 2
y'= 1/2 y - 1 + 2
y' = 1/2 y + 1 ...}kali 2
2y' = y + 2
y = 2 - 2y' ....(ii)


(i) dan (ii) substitusikan ke 
(y) = (x)² + 3
(2 - 2y') = (1- 2x')² + 2
2 - 2y' = 1 - 4x' + 4x'² + 2
- 2y' =  4x'² - 4x'  + 1
y' = -1/2 (4x'² -  4x' +1)
atau
y = -1/2 (2x  - 1)²

Persamaan garis : [y = x² + 3]

dilatasi [ P(a,b), D (k) ] = {P(1,2 ) , k = 1/2}

x' = k (x - a) + a
x' = 1/2 (x - 1) + 1
x' = 1/2 x - 1/2 + 1
x' = 1/2 x + 1/2 (×2)
2x' = x + 1
x = 1 - 2x' ----> (i)

y' = k(y-b) + b
y' = 1/2 (y - 2) + 2
y'= 1/2 y - 1 + 2
y' = 1/2 y + 1 (×2)
2y' = y + 2
y = 2 - 2y' ----> (ii)


substitusikan :
(y) = (x)² + 3
(2 - 2y') = (1- 2x')² + 2
2 - 2y' = 1 - 4x' + 4x'² + 2
- 2y' = 4x'² - 4x' + 1
y' = -1/2 (4x'² - 4x' +1)
atau
y = -1/2 (2x - 1)²

maaf kalo salah

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