#1**+10 **

There was another answer here before and it was correct. What happened to it? This forum is full of ghosts!

This question is making me think.

Consider y=x^{2} (I am only going to look at it for x>=0)

The inverse function is x=y^{2} which simpifies to y=sqrt(x)

Now thie inverse function is the reflection of the function in the line y=x

NOW Both the original fuction y=x^{2} and its inverse y=sqrt(x) have been dropped by 2 units.

So it stands to reason that the line of reflection has also been dropped by 2 units.

So the line of reflection has become y=x-2

SO If the graph y=x^{2}-2 is reflected about the line y=x-2 it is transformed into y=sqrt(x)-2 where x>=0

There you go and the graph backs this logic up.

Melody
Jun 25, 2014

#1**+10 **

Best Answer

There was another answer here before and it was correct. What happened to it? This forum is full of ghosts!

This question is making me think.

Consider y=x^{2} (I am only going to look at it for x>=0)

The inverse function is x=y^{2} which simpifies to y=sqrt(x)

Now thie inverse function is the reflection of the function in the line y=x

NOW Both the original fuction y=x^{2} and its inverse y=sqrt(x) have been dropped by 2 units.

So it stands to reason that the line of reflection has also been dropped by 2 units.

So the line of reflection has become y=x-2

SO If the graph y=x^{2}-2 is reflected about the line y=x-2 it is transformed into y=sqrt(x)-2 where x>=0

There you go and the graph backs this logic up.

Melody
Jun 25, 2014