Given reflection of graph, write the equation

hope4ever

New member
Joined
Sep 12, 2006
Messages
2
I am a little bit confused about one math question. Can you please help me out? There are two parts; I already finished part (a), but I am stuck on part (b).

Part a: Draw the reflection of f(x) = (square root X-2) in the line y = x.

I understand the above, but when it comes to part (b), I am totally clueless.

Part b: Write the equation of the reflection in the form y = g(x)

I believe the equation is (square root x+2) + 2, but x => 2 and y => 0. But what do I do with this information?

Thanks for your help!
 

steve_b

Junior Member
Joined
May 2, 2004
Messages
78
Part a: Draw the reflection of f(x) = (square root X-2) in the line y = x.

I understand the above, but when it comes to part (b), I am totally clueless.

Part b: Write the equation of the reflection in the form y = g(x)

I believe the equation is (square root x+2) + 2, but x => 2 and y => 0. But what do I do with this information?

=============

The reflection of a graph around the y=x line will be the inverse function. So you are given:

y = sqrt(x-2)

We get two pieces of information from this function. First, x must be greater than or equal to 2, or the radicand will be negative (can't have that with real numbers). Second, y must be greater than or equal to 0 since it is a square root, and principal square roots are defined as positive (or 0).

So we have to find the inverse function of y = sqrt(x-2)

First step, interchange the variables to get:

x = sqrt(y-2)

Now solve for y by first squaring both sides:

x^2 = y - 2

y = x^2 + 2

This function exists for all x, but the original function existed for x > 2, so we have to limit the domain of this new function to x >0 or it won't be a true reflection. In addition, the range of the new function is y >2, since 2 is it's smallest value. If you graph both functions you will see this.

Hope that helps...

Steve
 
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