If y = [(4 - x)]^2, what is the smallest possible y value?
If the vertex had been below or above the x-axis, setting the equation equal to zero and solving will only find you the zeros (x-intercepts), not the minimum point.
To find the x-coordinate of the minimum/maximum point use
Then, substitute that into your original equation to find the y-coordinate. Thus, the minimum/maximum point would be
In any case, the whole point was not to find the x for which y= 0 but the fact that y can be 0 which should have been obvious.
0 = 4 - x
-4 = -x
-4/-1 = x
4 = x
This is where the point (4, 0) comes from indicating that the smallest possible y value is 0, right?