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 $\binom{n}{k}=\frac{n!}{k!(n-k)!},$ $$\sum\_{n=k}^\infty\binom{n}{k}x^{n}=\frac{x^k}{(1-x)^{k+1}}.$$
$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$$
\begin{array}{rclr}
\ds{\color{#f00}{\sum\_{n = k}^{\infty}{n \choose k}x^{n}}} & 
\ds{\stackrel{n\ \mapsto\ n + k}{=}} &\!\!\!\!\!
\ds{\sum\_{n = 0}^{\infty}{n + k \choose k}x^{n + k}}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{x^{k}\sum\_{n = 0}^{\infty}{n + k \choose n}x^{n}} &\pars{~Binomial\ Symmetry~}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{x^{k}\sum\_{n = 0}^{\infty}{\bracks{-n - k} + n - 1 \choose n}\pars{-1}^{n}x^{n}} &
\qquad\pars{~Negating\ Property~}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{x^{k}\sum\_{n = 0}^{\infty}{- k - 1 \choose n}\pars{-x}^{n} =
x^{k}\bracks{1 + \pars{-x}}^{-k - 1}} & \pars{~Binomial\ Theorem~}
\\[5mm] & \ds{=} &\!\!\!\!\!
\ds{\color{#f00}{x^{k} \over \pars{1 - x}^{k + 1}}} &
\end{array}
$$