Question

Fungsi kuadrat yang grafiknya memiliki titik balik P(3,9) dan melalui titik (5,13) adalah
a. f(x)=(x+3)^2+5
b. f(x)=(x+3)^2+9
c. f(x)=(x-3)^2+5
d. f(x)=(x-3)^2+9
e. f(x)=(x-3)^2+13
jawab dengan caranya juga ya guys.

Answer 1

Cara pertama
P(3, 9) -> Xpuncak = 3, Ypuncak = 9
A(5,13) -> x = 5, y = 13

Xpuncak = - b/2a
3 = - b/2a
-b = 6a
b = - 6a

Ypuncak = 9
(b^2 - 4ac)/-4a = 9
(-6a)^2 - 4ac = - 36a
36a^2 - 4ac = - 36a
4a(9a-c) = - 36a
9a - c = - 9
c = 9a + 9
F(x)=ax^2 + bx + c
F(5)= 25a + 5b + c
13 = 25a + 5(-6a) + 9a + 9
13 = 25a - 30a + 9a + 9
4 = 4a
a = 1

B = - 6a = - 6
C= 9a + 9 = 9 + 9 = 18

F(x) = ax^2 + bx + c
F(x) = x^2 - 6x + 18
F(x) = x^2 - 6x + (-6/2)^2 + 18 - (-6/2)^2
F(x) = (x - 6/2)^2 + 18 - 3^2
F(x) = (x-3)^2 + 18 - 9
F(x) = (x-3)^2 + 9

Cara kedua
P(3,9) -> h = 3, k = 9
A(5,13) -> x = 5, y = 13

(x-h)^2 = 4p(y-k)
(5-3)^2 = 4p(13-9)
2^2 = 4p(4)
4 = 16p
p = 1/4

(x-h)^2 = 4p(y-k)
(x-3)^2 = 4(1/4)(y-9)
(x-3)^2 = y - 9
y = (x-3)^2 + 9
F(x) = (x-3)^2 + 9