Respuesta :
[tex]\bf 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}}\\\\
f(x)=&{{ A}} \left|{{ B }}x+{{ C}} \right|+{{ D}}
\\\\
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[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]
with that template in mind, let's see,
[tex]\bf f(x)=|x| \implies \begin{array}{lllccll} f(x)=&1|&1x&+0|&+0\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}[/tex]
so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.
the smaller A is, the wider it opens, the larger it is, the more it shrinks.
[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]
with that template in mind, let's see,
[tex]\bf f(x)=|x| \implies \begin{array}{lllccll} f(x)=&1|&1x&+0|&+0\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}[/tex]
so, to shift it to the right by 2 units, simply set C = 2.
to stretch it by 2, set A = 1/2.
the smaller A is, the wider it opens, the larger it is, the more it shrinks.
The function is: [tex]g(x) = 2|x - 2|[/tex]
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This question is solved using translation and stretching concepts.
- Translating a function f(x) a units to the right is the same as finding f(x - a).
- Vertically stretching a function f(x) a units is the same as finding a*f(x), that is, multiplying f(x) by a.
-------------------------
Original function:
The original function is:
[tex]f(x) = |x|[/tex]
-------------------------
Translation 2 units to the right
f(x - 2), so:
[tex]f(x - 2) = |x - 2|[/tex]
-------------------------
Horizontal stretch by a factor of 2
Multiplying by 2, so:
[tex]g(x) = 2f(x - 2) = 2|x - 2|[/tex]
In the graph at the end of this question, there original function f(x), in red, and the function g, in blue, are given.
A similar question is given at https://brainly.com/question/17072859
