Andrew L. answered • 05/25/14

Awesome Algebra!

Hi Trisha,

I wish I was able to attach some pictures to the post to illustrate, but let me go ahead and explain the best I can with just writing.

## Shifting...what's that?

Whenever we want to shift a function, in any direction, our goal is to manipulate the variable of the direction we are shifting in. In this case...we want to shift along the 'x' axis (to the right) so we will be messing with the 'x' term of our function.

Not too bad so far!

## So how do we shift anyway??

Well...if we want to move a function to the right we want to subtract 9 units from the 'x' term. Why? you ask. Let's take a look:

If we shift to the right, we would like f(x)--our original function, to correspond with f(x-9)--our shifted function.

For example:

the y value at x=0 on our original function should exist at x=9 on our shifted function. You can visualize this by making a graph and placing a dot at the point (0,0). Move that point 9 places over to the right. The point is now at (9,0). To get this to happen mathematically we want to be able to plug x=9 into our shifted function and get the value that once existed at x=0 in our original function. This is why we want our shifted function to be a copy of our original function except subtracting 9 from the 'x' that is inputted. f(x-9) will shift a function the right 9 places!!

**Get on with it...what's the answer already:**

**original function:**f(x) = (x+8)^2

**Shifted function :**f(x-9) = (x-9 +8) ^2 **Everything in orange is when we plug in "x-9" for "x"

**Final shifted function: f(x-9) = (x-1)^2 ** *This is our shifted function!

## Final thoughts from Andrew L. :

I strongly suggest you graph both of these functions on a graphing calculator (on your own) and try to understand why this shift occurred... Try to make sense regarding how subtracting 9 from the 'x' variable shifted our function 9 spaces to the right.

If you have any questions or would like additional clarification, PLEASE PLEASE PLEASE email me or comment here and I'll be happy to go into more detail regarding algebraic manipulations. Good luck!!

Best Regards,

Andrew L.