Respuesta :
[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
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\begin{array}{llll}
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 \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the y-axis} \\\\[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. 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}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}}[/tex]
that said, let's say then, we have f(x)
2f(x), will make A = 2, and therefore, will "shrink" the graph vertically by a factor of twice as much from f(x).
f(-x) or f(-1x) will make B = -1, therefore, will simply flip the graph over the y-axis, rotation over the y-axis, just a mirror image.
f(x) - 1 will make D = -1, that means, a downward vertical shift of 1 unit, so the graph will drop by 1.
2f(-x) -1, shrunk by twice, mirror over the y-axis, and dropped by 1.
[tex]\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\ \left. \qquad \right. \textit{reflection over the y-axis} \\\\[/tex]
[tex]\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. 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}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}}[/tex]
that said, let's say then, we have f(x)
2f(x), will make A = 2, and therefore, will "shrink" the graph vertically by a factor of twice as much from f(x).
f(-x) or f(-1x) will make B = -1, therefore, will simply flip the graph over the y-axis, rotation over the y-axis, just a mirror image.
f(x) - 1 will make D = -1, that means, a downward vertical shift of 1 unit, so the graph will drop by 1.
2f(-x) -1, shrunk by twice, mirror over the y-axis, and dropped by 1.