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# Graph

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How would I graph thisâ€‹

$$f(x)=\sqrt[2]{x+1}-2$$

Jan 18, 2018

#4
+99117
+2

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 :)

Jan 18, 2018

#1
+98005
+1

This is the square root graph   shifted to the  left by 1 unit and down 2 units

See  both here :

https://www.desmos.com/calculator/c2df9wir6j

Jan 18, 2018
#2
+1

I was wondering how you would graph this by hand

Jan 18, 2018
#3
+98005
+2

Ok....let's pick some points  and put them into the function to get y.....

To save time.... i'm going to pick some points for x  and have WolframAlpha  generate the associated y values

You can plot these for yourself to get a rough idea of the graph

sqrt(x + 1) - 2 where

x = {-1, -0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3}

y  =  {-2, -1.29289, -1, -0.775255, sqrt(2) - 2, -0.418861, sqrt(3) - 2, -0.129171, 0}

Jan 18, 2018
#4
+99117
+2

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 :)

Melody Jan 18, 2018