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Correct me if I am wrong, but I think that, as a general rule:

  • When the input of a function is diminished, that is when $f(x)$ becomes $f( x-n )$ with $(n>0)$ , then the graph is moved to the right.

  • When the input of a function is increased, that is when $f(x)$ becomes $f( x+n )$ with $( n>0)$, then the graph is moved to the left.

It took me some time to understand this, but now it seems to me that I come across an exception, with $f(x) = \sqrt{ - x -1}$ which moves 1 unit to the left relatively to $f(x)= \sqrt{ - x}$.

Could someone explain to me what I've missed?


NOTE : The explanations given below make me realize what was wrong in my question.

I thought that the case I indicate was an exception to the general rule :

" f(x-n) ( with n positive) produces a horizontal shift to the right".

BUT , actually, f(x) = sqrt( -x - 1 ) is NOT a case that falls under this rule.

In order to have a case falling under the rule, I should have treated ( x-1) as the new input, in order to get : f(x-1) = sqrt [ - ( x-1) ] = sqrt (-x +1).

The new function f ( now correctly defined) perfectly obeys the rule: with a graphing calculator, I can see that the graph moves 1 unit to the right.

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You have $f(x)=\sqrt{-x}$, and by "subtract $ 1 $ from $ -x $" what you're doing is calculate $f(x+1)$, which moves the original graph of $ f (x) $ one unit to the left. Look $$f(x+1)=\sqrt{-(x+1)}=\sqrt{-x-1}$$ What you want to do is compute $ f (x-1) $, that is, $ \sqrt{-x + 1} $ so that the graph moves one unit to the right.

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    $\begingroup$ Arziff00.Thanks for this clear answer! Apparently, I still have some progresses to make in precalculus-algebra. $\endgroup$
    – user654868
    Jul 17 '19 at 17:03
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When the INPUT of a function is increased/decreased....

Notice you are subracting the $1$ somewhere in the output.

If $f(x) = \sqrt{-x}$ then $f(x+1) = \sqrt{-(x + 1)} = \sqrt{-x -1}$ and $f(x-1) = \sqrt{-(x-1)} = \sqrt{-x+1}$.

So $-1$ in the output does not mean a $-1$ in the input. In fact it absolution means a $+1$ in the input.

....

Here's another "paradox" Let $f(x) = 4x$ and $f(x)$ passes through $(0,0)$ but $4x + 1$ which is $+1$ should be a of $1$ to the left, passes through $(0,-\frac 14)$ so the graph is only shifted $\frac 14$ to the left. How did a $1$ become $\frac 14$.

Again a $+1$ in the ouput doesn't mean a $+1$ in the input. Noting $f(x+\frac 14) = 4(x +\frac 14) =4x + 1$ , a s $+1$ in the output means a $+\frac 14$ in the input.

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    $\begingroup$ This is very important point. $\endgroup$
    – NoChance
    Jul 17 '19 at 23:24
  • $\begingroup$ @fleablood. Very helpful! Thanks. The point is subtle. $\endgroup$
    – user654868
    Jul 18 '19 at 15:41

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