[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}[/tex]
[tex]\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}[/tex]
[tex]\bf \begin{array}{llll}
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}[/tex]
so hmm if you notice, your f(x) there, looks just like g(x), but is "shifted horizontally" by about 9 units to the right
that simply means, based on that template above, that C/B = -9, and we can simply make B = 1 and C = -9, and we get C/B = -9/1 which is -9
thus [tex]\bf \begin{array}{llcll}
f(x)=g(&1x&-9)\\
&\uparrow &\uparrow \\
&B&C
\end{array}\iff f(x)=g(x-9)[/tex]
