To shift a graph horizontally, you include the shift amount WITH x and perform whatever is being done otherwise just to x, to x and the shift amount.
Let's look at an example:
f(x) = x^2 + 4 ................. and f(x) = (x-3)^2 + 4.
You can see what I mean by including the shift amount WITH x. The 'square' function acts on the entire (x-3) term. This will cause the graph to shift 3 units to the RIGHT. This may seem somewhat counter-intuitive, but it is correct.
In this example, x-3 shifts 3 units right... if it were x+3, it would shift 3 units left. It may be easier to remember this by analyzing the x and shift amount, letting that small term equal to 0, and then solving for x. Like this:
x - 3 = 0
x = 3
OR
x + 3 = 0
x = (-3)
That shows you how far over, and in what direction, the new x values are!
I hope these postings on graph manipulations are helpful. Please leave a comment if you are unsure of something, as there are likely many other students who might be wondering the exact same thing!
2 comments:
I see what you are doing with the f(x+3) and f(x-3) examples as they = zero.
I can remember that within parentheses we shit x in the counterintuitive direction.
That being said though, why isn't this more straigtforward "looking?"
Wouldn't it have been simpler simply to not confuse people by simply having the meaning of x-3 within parentheses mean what it appears to mean?
Why do we have to torture ourselves like that.
Is there a rational reason to make this appear as it is not?
Silvia - 16 - Junior to be
Unfortunately, I'm not the one making the rules here. :) I agree that it can be confusing and counter-intuitive, but that's just the way it works. These concepts weren't devised to be 'straightforward looking,' but rather to be functional and correct. If we were to just make f(x-3) be shifted to the left instead of the right, nothing else would make sense! Overall, small and confusing points add up to make the big picture work out in the end. It just takes practice to get it. :)
Post a Comment