[tex]\bf ~~~~~~~~~~~~\textit{function transformations}
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% templates
f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}}
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~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
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f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
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f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
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f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
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--------------------[/tex]
[tex]\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
~~~~~~if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
~~~~~~if\ {{ D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}[/tex]
with that template in mind, let's check,
[tex]\bf f(x)=x^2\implies f(x)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+0})^2\stackrel{D}{+0}
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g(x)=-(-x)^2\implies g(x)=\stackrel{A}{-1}(\stackrel{B}{-1}x\stackrel{C}{+0})^2\stackrel{D}{+0}[/tex]
A became -1, reflection over the x-axis
B became -1, reflection over the y-axis
so g(x) is just f(x) upside-down and right-side to the left, check the picture below.
