What is the vertex of #y=sqrt(x+2)#?

1 Answer
Jun 25, 2015

I do not think this function has a vertex (considered as the highes or lowest point as in a parabola).


The square root, such as this one, has a graph that looks like a horizontal half parabola.
If you mean the hypotetical vertex of the complete parabola then you have that its coordinates are #x=-2, y=0# but I am not sure it can be considered as a proper vertex:
The graph looks like this:
graph{sqrt(x+2) [-10, 10, -5, 5]}
As you can see you have only half parabola!