Answer:
[tex]1.\\\text{When $x=-2$, we have $y = 3(-2) + 4 = -6 + 4 = -2$}\\\text{When $x=-1$, we have $y = 3(-1) + 4 = -3 + 4 = 1$}\\\text{When $x=0$, we have $y = 3(0) + 4 = 0 + 4 = 4$}\\\text{When $x=1$, we have $y = 3(1) + 4 = 3 + 4 = 7$}\\\text{When $x=2$, we have $y = 3(2) + 4 = 6 + 4 = 10$}[/tex]
[tex]2.\\\text{When $x=-2$, we have $y = (-2)^2 - 3 = 4 - 3 = 1$}\\\text{When $x=-1$, we have $y = (-1)^2 - 3 = 1 - 3 = -2$}\\\text{When $x=0$, we have $y = (0)^2 - 3 = 0 - 3 = -3$}\\\text{When $x=1$, we have $y = (1)^2 - 3 = 1 - 3 = -2$}\\\text{When $x=2$, we have $y = (2)^2 - 3 = 4 - 3 = 1$}[/tex]
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Answer:
[tex]1.\\\text{When $x=-2$, we have $y = 3(-2) + 4 = -6 + 4 = -2$}\\\text{When $x=-1$, we have $y = 3(-1) + 4 = -3 + 4 = 1$}\\\text{When $x=0$, we have $y = 3(0) + 4 = 0 + 4 = 4$}\\\text{When $x=1$, we have $y = 3(1) + 4 = 3 + 4 = 7$}\\\text{When $x=2$, we have $y = 3(2) + 4 = 6 + 4 = 10$}[/tex]
[tex]2.\\\text{When $x=-2$, we have $y = (-2)^2 - 3 = 4 - 3 = 1$}\\\text{When $x=-1$, we have $y = (-1)^2 - 3 = 1 - 3 = -2$}\\\text{When $x=0$, we have $y = (0)^2 - 3 = 0 - 3 = -3$}\\\text{When $x=1$, we have $y = (1)^2 - 3 = 1 - 3 = -2$}\\\text{When $x=2$, we have $y = (2)^2 - 3 = 4 - 3 = 1$}[/tex]