Write an equation that satisfies the relationship shown in the table of value.

x y
-2 0
0 4
2 8
5 14
8 20

To find an equation that satisfies the relationship shown in the table of values, we need to look for any patterns or relationships between the x-values and y-values. In this case, we can observe that as the x-values increase, the y-values also increase, but by different amounts.

Let's examine the differences between consecutive y-values:
- For x = -2 to x = 0, the y-value increases by 4 (from 0 to 4).
- For x = 0 to x = 2, the y-value increases by 4 (from 4 to 8).
- For x = 2 to x = 5, the y-value increases by 6 (from 8 to 14).
- For x = 5 to x = 8, the y-value increases by 6 (from 14 to 20).

From this, we can see that the differences between consecutive y-values are not constant. Therefore, we can conclude that the relationship is not linear.

To find an equation that satisfies the given relationship, we need to look for a pattern in the differences. In this case, the differences are increasing by 2 each time. Based on this pattern, we can assume that the relationship might be quadratic.

To write an equation for a quadratic relationship, we use the general form:

y = ax^2 + bx + c

where a, b, and c are constants to be determined.

Let's substitute the given x-values and y-values into the equation and solve for the constants:

For x = -2, y = 0:
0 = a(-2)^2 + b(-2) + c (Equation 1)

For x = 0, y = 4:
4 = a(0)^2 + b(0) + c (Equation 2)

For x = 2, y = 8:
8 = a(2)^2 + b(2) + c (Equation 3)

Simplifying Equations 1 and 2, we get:
0 = 4a - 2b + c (Equation 4)
4 = c (Equation 5)

Substituting Equation 5 into Equation 1, we have:
0 = 4a - 2b + 4 (Equation 6)

Simplifying Equation 6, we get:
-4 = 4a - 2b (Equation 7)

Substituting Equation 5 into Equation 3, we have:
8 = 4a + 2b + 4 (Equation 8)

Simplifying Equation 8, we get:
4 = 4a + 2b (Equation 9)

Subtracting Equation 7 from Equation 9, we eliminate the "b" term:
4 - (-4) = 4a + 2b - (4a - 2b)
8 = 4b

So, b = 2.

Substituting this value of "b" into Equation 9, we can solve for "a":
4 = 4a + 2(2)
4 = 4a + 4
4a = 0
a = 0

Now that we have found the values of "a" and "b", we can substitute them back into Equation 5 to find the constant "c":
4 = c

Therefore, the equation that satisfies the relationship shown in the table of values is:

y = 0x^2 + 2x + 4

or simply:

y = 2x + 4