suatu fungsi rumuskan dengan f(x)= ax+b jika f(-2) = 14 dan f(3) = -1 nilai f(7) -f(10) adalah
Matematika
ZACK133
Pertanyaan
suatu fungsi rumuskan dengan f(x)= ax+b jika f(-2) = 14 dan f(3) = -1 nilai f(7) -f(10) adalah
2 Jawaban
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1. Jawaban superpaw99
f(-2) = 14 -> f(-2) = -2a + b -> 14 = -2a + b
f(3) = -1 -> f(3) = 3a + b -> -1 = 3a +b
---------------------- -
15 = -5a
a = -3
-1 = 3(-3) + b --> -1 = -9 + b --> b = 7
f(7) = 7a + b --> 7(-3) + 7 = -14
f(10) = 10a + b --> 10(-3) + 7 = -23
f(7) - f(10) = -14 - (-23) = 9 -
2. Jawaban Nik001
f(x) = ax + b
f(-2) = -2a + b = 14
f(3) = 3a + b = -1
-------------------------- -
-5a = 15
a = 15/-5
a = -3
3a + b = -1
3(-3) + b = -1
-9 + b = -1
b = -1 + 9
b = 8
f(x) = ax + b
f(7) = -3(7) + 8
= -21 + 8
= -13
f(10) = -3(10) + 8
= -30 + 8
= -22
f(7) - f(10) = -13 - (-22)
= -13 + 22
= 9