Mmm Lets see.
\(f(x)=\sqrt[2]{x+1}-2\)
this is just
\(f(x)=\sqrt[2]{x+1}\) dropped 2 units
so I will look at this one:
\(y=\sqrt[2]{x+1}\\ \qquad \text{I first note that }y\ge0\qquad and \qquad x\ge-1\\ y^2=x+1\\ x=y^2-1\\ \text{This is a sideways parabola. It will open up on the positive side }.\\ \text{The parent function for this one is }x=y^2\\ \text{The -1 means that it is moved 1 unit to the left }\)
SO what do we have - we have to go backwards now:
Graph \(x=y^2\) which is a sideways parabola with vertex (0,0)
Now move it 1 unit to the left.
Now only include the bits where \( x\ge-1 \quad and \quad y\ge0\)
NOTE: This is now the same as \(y=\sqrt{x+1}\)
Now drop it two units
and then you should have it. \(y=\sqrt{x+1}-2\)
I will show you how this graph grows from the parent (or is that grandparent) function:
Here is how the graph is built up.
You can play withh this, if you press the coloured circles on the left your graph will diappear or reappear, so you can see how I built the graph :)
https://www.desmos.com/calculator/0fgnsho6gv
this is just the pic to make the post pretty, but you should to to the linked site :)