If y = [(4 - x)]^2, what is the smallest possible y value?
What you did was solve for the x-intercepts. This is where y=0. This is the smallest value for y only when the parabola is tangent to the x-axis, as in this case. The vertex is on the x-axis at (4, 0)
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
from
Then, substitute that into your original equation to find the y-coordinate. Thus, the minimum/maximum point would be
Why in the world would you multiply out an expression that was already factored and then factor it again?
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?