# Why does subtracting 1 from $-x$ in $y= \sqrt{-x}$, getting $y = \sqrt {-x -1 }$, result in a LEFT horizontal shift?

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.

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.

• Arziff00.Thanks for this clear answer! Apparently, I still have some progresses to make in precalculus-algebra.
– user654868
Jul 17 '19 at 17:03

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.

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