ada yg bisa bantu jawab no 1?

fungsi dan relasi

1. diket: f(x)=x²-3x+2

dit:
d= {-2,-1,0,1,2}

jawab:

f(x)= x²-3x+2
f(-2)=(-2)²-3(-2)+2=4+6+2=12
f(-1)= (-1)¹-3(-1)+2=1+3+2=6
f(0)=(0)²-3(0)+2=2
f(1)= (1)¹-3(1)+2=1-3+2=0
f(2)=(2)²-3(2)+2=4-6+2=0

jadi range adalah {12,6,2,0,0}

semoga membantu jawaban ini

f(x) = x^2 - 3x + 2

D = {-2 , -1 , 0 , 1 , 2}

f(x) = x^2 - 3x + 2

f(-2) = (-2)^2 - 3(-2) + 2
f(-2) = 4 + 6 + 2
f(-2) = 12

f(-1) = (-1)^2 - 3(-1) + 2
f(-1) = 1 + 3 + 2
f(-1) = 6

f(0) = 0^2 - 3(0) + 2
f(0) = 0 - 0 + 2
f(0) = 2

f(1) = 1^2 - 3(1) + 2
f(1) = 1 - 3 + 2
f(1) = 0

f(2) = 2^2 - 3(2) + 2
f(2) = 4 - 6 + 2
f(2) = 0


Jadi range nya adalah R = {0 , 2 , 6 , 12}