Answer:
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)
The point (2,6) is not on the graph of the linear equation
The points on the linear equation are given as:
(-2,0) and (0,-6)
A linear equation is an equation that has a constant rate
The above definition means that, as the x values increases by 2, the y values reduces by 6
So, we have the following points on the linear equation
(-2,0), (0,-6), (2, -12)
Notice that (2,-12) and (2,6) are not the same.
Hence, the point is not on the graph of the linear equation
Read more about linear equations at:
https://brainly.com/question/14323743
