# Graph of the function in the form: sqrt(-x+a) through manipulation of the function sqrt(x)?

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?

• $f(-x+a)=f(-(x-a))=g(x-a)$ 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. – zoli Oct 31 '15 at 23:53
• Ok, so reflect f across y, then shift the reflection to the right. So this means that f(x) = sqrt(x-a)? – RiddleMeThis Nov 1 '15 at 4:45

• 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 $(1,\infty)$.
• @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: $g(x)=f(-x)$. – zoli Nov 1 '15 at 21:14