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In class we learned that certain simple functions $(x, x^2, \sqrt x)$ etc. can be manipulated to easily find the graphs of more complicated versions of these original functions... $f(x+1) \Longrightarrow$ shift to the left/ $f(x-1)\Longrightarrow$ shift to right, etc.

However, I'm having trouble applying that to the function: $f(x) = \sqrt{(-x+a)}.$
So far I understand this much: $\sqrt x \Longrightarrow \sqrt{(-x)}$ results in reflection across $y-$axis.
$\sqrt x \Longrightarrow \sqrt{(x\pm a)}$ results in shift of original graph a units to the left/right, respectively.

But when you apply the principles to $\sqrt{(-x+a)}$ or $\sqrt{(-x-a)},$ it doesn't respond appropriately. The graph I get for $\sqrt{(-x+a)}$ is the graph of $\sqrt{(-x)}$ shifted a units to the right , not the left.
And the opposite goes for $\sqrt{(-x-a)}.$
Why is this? Can normal manipulation of the graph not be applied to functions of this form?

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    $\begingroup$ $f(-x+a)=f(-(x-a))=g(x-a)$ where $g(x)=f(-x)$. So, one has to shift $g$ to the right and $g$ is the reflection of $f$ over $y$. The correct order: (1) reflecting $f$ (2) then shifting the reflection to the right. $\endgroup$ – zoli Oct 31 '15 at 23:53
  • $\begingroup$ Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)? $\endgroup$ – RiddleMeThis Nov 1 '15 at 4:45
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I would like to illustrate my comments in the following figure. I hope that this will explain everything.

enter image description here

  • the dark blue line is the graph of $\color{blue}{\sqrt{(x)}}$ -- note that $\sqrt{(x)}$ is not defined on $(-\infty,0)$.
  • the purple line is the graph of $\color{purple}{g(x)=\sqrt{(-x)}}$ -- note that this function is not defined on $(0,\infty)$.
  • The green line is the graph of $\color{green}{g(x-1)=\sqrt{-(x-1)}=\sqrt{-x+1}}$ -- note that this function is not defined if $-(x-1)<0$ or if $x>1$, that is over the interval $(1,\infty)$.
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  • $\begingroup$ ok. The graphs of sqrt(x) and sqrt(-x) make sense to me because they are each reflections of each other across the y-axis. I also understand how the graph of sqrt(x+1) would look like sqrt(x) shifted 1 unit to the left on x-axis. Then you negate every x in that function, by doing f(-x) and this becomes a reflection across y-axis to yield what you drew with the green line. The thing I was doing wrong was that you can't replace x with -x+1. You can only replace x in sqrt(x+1) with -x. So sqrt(-x+1) is a reflection of sqrt(x+1). Is this correct? $\endgroup$ – RiddleMeThis Nov 1 '15 at 16:39
  • $\begingroup$ @RiddleMeThis: The rule (my rule) is that if you have a function then you have only three legal transformations $f(x+a)$, $f(x-a)$, and $f(-x)$. So, $f(-x\pm a)$ is not legal. You can make it legal via the following operation $f(-x\pm a)=f(-(x\mp a))$ but then you've created another function: $g(x)=f(-x)$. $\endgroup$ – zoli Nov 1 '15 at 21:14

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