Why does shifting a function to the right require a smaller input variable?

AI Thread Summary
Shifting a function horizontally involves altering the input variable before applying the function. For example, in f(x-1), the input x is decreased by 1, which means to achieve the same output, the graph shifts to the right. This is because a smaller input results in the same output as a larger input in the original function. The discussion clarifies that when shifting right, the input must be smaller, while shifting left would require a larger input. Understanding this concept is crucial for accurately interpreting function transformations.
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Homework Statement


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?


Homework Equations



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

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?
 
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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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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