Kurva fungsi kuadrat melalui titik (3,2) dan berpuncak pada titik (1,-3). Persamaan kurvannya adalah…

Jawab:

P(xp,yp) = (1,-3)

melalui titik (x,y) = (3,2)

y = a (x – xp)² + yp

2 = a (3-1)²  + (-3)

4a = 5

a = 5/4

substitusikan a

y = a (x – xp)² + yp

y = 5/4 (x-1)² – 3

y = 5/4 (x² -2x+1) -3

y = 5/4x² – 5/2 x + 5/4 – 3

y = 5/4x² – 5/2 x – 7/4

correct me if im wrong

Penjelasan dengan langkah-langkah:

TP = (1 , -3)

xp = 1 , yp = -3

T = (3 , 2)

x = 3 , y = 2

y = a(x – xp)^2 + yp

melalui titik (3 , 2) :

y = a(x – xp)^2 + yp

<=> 2 = a(3 – 1)^2 – 3

<=> 2 = a(4) – 3

<=> 2 = 4a – 3

<=> 4a = 2 + 3

<=> 4a = 5

<=> a = 5/4

y = 5/4(x – 1)^2 – 3

y = 5/4(x^2 – 2x + 1) – 3

y = 5/4 x^2 – 5/2 x + 5/4 – 3

y = 5/4 x^2 – 5/2 x + 5/4 – 12/4

y = 5/4 x^2 – 5/2 x – 7/4