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

Question

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?

Answer 1

I would like to illustrate my comments in the following figure. I hope that this will explain everything.

• 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 $(1,\infty)$.