Horizontal shift of function

  • Thread starter greenneub
  • Start date
1. The problem statement, all variables and given/known data
Could someone please explain, in very simple words, the horizontal shift of a graph? I've used the search button, textbooks, and google and I'm just not comprehending this. Why does the graph f(x-1) shift f(x) 1 unit to the right and not left?


2. Relevant equations

f(x) = x Let's keep it simple.

3. The attempt at a solution
If I take some ordered pairs, say (0,0), (1, 1), (2,2) as points on f(x), then try to create the new ordered pairs with f(x-1), shouldn't they be (-1, 0), (0, 1) (1,2) and the f(x-1) graph be moved 1 unit to the left?
 

HallsofIvy

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Because the first thing you do in evaluating f(x- 1) is subtract 1 from x. In other words, if g(x)= f(x-1), g(1)= f(0), g(2)= f(1), etc. The point on the original graph that was at 0 is not at 1, that was at 1 is now at 2, etc.
 
Take the graph of y = x^2. It's vertex is at the origin. Now the graph of (x-1)^2 is shifted to the right because to make y = 0, you have to add 1 etc.
 
When you shift a function up or down, you have already performed the function f(x) for all the points and you are simply moving the curve up or down the y-axis by adding or subtracting a constant to the result.

When shifting a function left or right, you are shifting the input variable x prior to performing the function. So, for example, if the curve has a positive slope and you are trying to shift it right, then the input variable needs to be some constant SMALLER than the current input so that the resulting curve shifts to the right. If you made the input variable larger, you would have a result for each point that is further up and out on the curve, resulting in a shift of the curve to the left, which is the opposite of what you wanted.
 

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