The points (-2,0) and (0,-6) are each on the graph of a linear equation. is (2,6) also
on the graph of this linear equation? explain your reasoning.

No, it does not. See below.

Step-by-step explanation:

Lets find out the linear equation that passes trough (-2, 0) and (0, -6).

We know every linear equation has the form: y = mx + b

Where m is the slope on the curve and b the independent term.

We know that, given 2 points (x1,y1) and (x2,y2) we can find the slope m as:

m = (y2-y1)/(x2-x1)

In our case lets replace (x1,y1) and (x2,y2) by (-2, 0) and (0, -6) (notice it could be done in the inverse sense where (-2, 0) is (x2,y2) and (0, -6) is (x1,y1) ). So, our slope is:

m = [-6 - 0] / [0 - (-2)]

m = -6/2 = -3

So, we have a downward linear function with slope -3, this is:

y =  -3x + b

Now, for finding b just replace any of the 2 points given in the equation. Lets replace (-2, 0):

0 = -3(-2) + b

0 = 6 + b

Subtracting 6 in both sides:

-6 = b

So, our independent term is -6 and the function is:

y = -3x - 6

Now lets see if this linear eqution passes trough (2,6). If it does, we can replace the values on the equation. Replacing x by 2:

y = -3(2) - 6 = = -6 - 6 = - 12

So, in our equation, we x is 2 y is -12, and not 6 as in the point (2,6). So, our equation does not passes trough (2,6)