Respuesta :
[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)=-3^{\stackrel{B}{2}x\stackrel{C}{-4}}\qquad \qquad g(x)=-3^{2x}\implies g(x)=-3^{\stackrel{B}{2}x\stackrel{C}{+0}}[/tex]
notice, g(x) has a horizontal shift of C/B or +0/2, or just 0, none.
while f(x) has a horizontal shift of C/B or -4/2, or -2, to the right.
so f(x) is really just g(x), but shifted horizontally over 2 units to the right.
[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)=-3^{\stackrel{B}{2}x\stackrel{C}{-4}}\qquad \qquad g(x)=-3^{2x}\implies g(x)=-3^{\stackrel{B}{2}x\stackrel{C}{+0}}[/tex]
notice, g(x) has a horizontal shift of C/B or +0/2, or just 0, none.
while f(x) has a horizontal shift of C/B or -4/2, or -2, to the right.
so f(x) is really just g(x), but shifted horizontally over 2 units to the right.
