Question: I want to sketch $|\sqrt{x}-5|$. Now I know what it is going to look like using basic graphing tools. 

My question is, is there an intuitive way to figure out what the graph looks like without making a table of values or using technology to see what the graph looks like?

I know $\sqrt{x}-5$ shifts the graph down five units, but I don't understand why the absolute value flips the graph.

Answer: 
For $|x|<1,$ we have $$\sum\_{n=0}^\infty x^n=\frac{1}{1-x}.$$ Now take $k$ derivatives on both sides: $$\sum\_{n=k}^\infty\frac{n!}{(n-k)!}x^{n-k}=\frac{k!}{(1-x)^{k+1}}.$$ Thus, $$\sum\_{n=k}^\infty\frac{n!}{(n-k)!k!}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$ Since $\binom{n}{k}=\frac{n!}{k!(n-k)!},$ $$\sum\_{n=k}^\infty\binom{n}{k}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$
$\sum\_{n=k}^{\infty}{{n}\choose{k}}x^n=\dfrac{x^k}{(1-x)^{k+1}}
$

Start with
$\sum\_{n=k}^{\infty}{{n}\choose{k}}x^n
=\sum\_{n=0}^{\infty}{{n}\choose{n+k}}x^{n+k}
=x^k\sum\_{n=0}^{\infty}{{n+k}\choose{k}}x^{n}
$

and note that
$(1-x)^{-m}
=\sum\_{k=0}^{\infty} \binom{-m}{k} x^k
$
and

$\begin{array}\\
\binom{-m}{k}
&=\dfrac{\prod\_{j=0}^{k-1}(-m-j)}{k!}\\
&=(-1)^k\dfrac{\prod\_{j=0}^{k-1}(m+j)}{k!}\\
&=(-1)^k\dfrac{\prod\_{j=0}^{k-1}(m+k-1-j)}{k!}\\
&=(-1)^k\dfrac{(m+k-1)!}{(m-1)!k!}\\
&=(-1)^k\binom{m+k-1}{k}\\
\end{array}
$