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a. f(1) =
-1 = 1 + p + q
-2 = p + q ... (1)
f(0) =
-5 = q
-2 = p + q
-2 = p - 5
3 = p
jadi nilai p = 3 dan q = -5
b. rumus fungsi =
c. f(5) =
6a. f(x) =
f(2) =
3 = 4 + 2a + b
-1 = 2a + b ...(1)
f(5) =
18 = 25 + 5a + b
-7 = 5a + b ... (2)
Eliminasi b dari persamaan (1) dan (2)
(2) - (1)
-6 = 3a
a = -2
-1 = 2*-2 + b
-1 = -4 + b
b = 3
b. Rumus fungsi:
c. f(-3) =
0 + 0 + q = -5 -
1 + p = 4
p = 3
1 + p + q = -1
1 + 3+ q = -1
q = -1 - 4
q = -5
b. f(x) = x² +3x - 5
c. f(x) = x² +3x - 5
= 5² + 3(5) - 5
= 25 + 15 - 5
= 35
6. a. 4 + 2a + b = 3
25 + 5a + b = 18 -
-21 - 3a = - 15
-3a = -15 + 21
-3a = 6
a = 6 : (-3)
a = -2
4 + 2a + b = 3
4 + 2(-2) + b = 3
0 + b = 3
b = 3 - 0
b = 3
b. g(x) = x² + 3x + 3
c. f(x) = x² + 3x + 3
f(-3)= (-3)² + 3(-3) + 3
= 9 - 9 + 3
= 3