[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]
now, with that template in mind, let's see
[tex]\bf \begin{array}{lllccll}
y=&1(&1x&+4)^{10}&-2\\
&\uparrow &\uparrow&\uparrow &\uparrow \\
&A&B&C&D
\end{array}\qquad from \qquad y=(x-0)^{10}\iff y=x^{10}
\\\\\\
\cfrac{C}{B}\implies \cfrac{4}{1}\implies +4\impliedby \textit{shifted to the right 4 units}
\\\\\\
D\implies -2\impliedby \textit{went down by 2 units}[/tex]
