[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}}\\\\
f(x)=&{{ A}} log\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\qquad[/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]
with that template in mind, let's see
[tex]\bf \begin{array}{lllcclll}
f(x)=&1log(&1x&+3)&+2\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}
\\\\\\
\textit{horizontal shift by }\cfrac{C}{B}\implies \cfrac{3}{1}\implies +3\textit{ units to the left}
\\\\\\
\textit{vertical shift by }D=+2\textit{ units upwards}[/tex]
