Question

Using a table of values to graph equations​

Answer 1

X, Process, Y

-6,
`(-6)/(3) +2`, 0

-3,
`(-3)/(3) +2`, 1

0,
`(0)/(3) +2`, 2

3,
`(3)/(3) +2`, 3

6,
`(6)/(3) +2`, 4

X = -6, -3, 0, 3, 6

Y = 0, 1, 2, 3, 4

Explanation:

Coordinates for graph are (-6,0), (-3,1), (0,2), (3,3), (6,4)

Graph is the image below.

Solve for Y

`(-6)/(3) +2` = 0

`(-3)/(3) +2` = 1

`(0)/(3) +2` = 2

`(3)/(3) +2` = 3

`(6)/(3) +2` = 4

Answer 2

Explanation:

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