[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 hmm g(x) to the right 8 units, meaning C/B is -8, so, you can just make C = -8 and B =1, -8/1 = -8
and down by 7 units, ok, that simply means D = -7
[tex]\bf \begin{array}{lllll}
g(x)=&1(&1x&+0)^2&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D\\
&\downarrow &\downarrow &\downarrow &\downarrow \\
g(x)=&1(&1x&-8)^2&-7
\end{array}\implies g(x)=1(1x-8)^2-7
\\\\\\
g(x)=(x-8)^2-7[/tex]
