Bantu jawab soal 1 dan 2 ​

1. y = |x + 3| limit (-7 ≤ x ≤ 2)

y = |-7 + 3| = |-4| = 4 → (-7,4)

y = |-6 + 3| = |-3| = 3 → (-6,3)

y = |-5 + 3| = |-2| = 2 → (-5,2)

y = |-4 + 3| = |-1| = 1 → (-4,1)

y = |-3 + 3| = 0 → (-3,0)

y = |-2 + 3| = |1| = 1 → (-2,1)

y = |-1 + 3| = |2| = 2 → (-1,2)

y = |0 + 3| = |3| = 3 → (0,3)

y = |1 + 3| = |4| = 4 → (1,4)

y = |2 + 3| = |5| = 5 → (2,5)

2. y = |2 – x| – 1 limit (-2 ≤ x ≤ 6)

y = |2 – (-2)| – 1 = |2 + 2| – 1 = 4 – 1 = 3 → (-2,3)

y = |2 – (-1)| – 1 = |2 + 1| – 1 = 3 – 1 = 2 → (-1,2)

y = |2 – 0| – 1 = |2| – 1 = 1 → (0,1)

y = |2 – 1| – 1 = |1| – 1 = 0 → (1,0)

y = |2 – 2| – 1 = 0 – 1 = -1 → (2,-1)

y = |2 – 3| – 1 = |-1| – 1 = 1 – 1 = 0 → (3,0)

y = |2 – 4| – 1 = |-2| – 1 = 2 – 1 = 1 → (4,1)

y = |2 – 5| – 1 = |-3| – 1 = 3 – 1 = 2 → (5,2)

y = |2 – 6| – 1 = |-4| – 1 = 4 – 1 = 3 → (6,3)