Respuesta :
[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 yours
[tex]\bf \begin{array}{llll} g(x)=&4ln(x)\implies &2(2)ln(x)\\ &\uparrow &\uparrow \\ &A&A \end{array}[/tex]
A is twice as large as in f(x), thus the graph shrinks twice as much in g(x)
[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 yours
[tex]\bf \begin{array}{llll} g(x)=&4ln(x)\implies &2(2)ln(x)\\ &\uparrow &\uparrow \\ &A&A \end{array}[/tex]
A is twice as large as in f(x), thus the graph shrinks twice as much in g(x)
Answer:
The graph of f(x) vertically stretch by factor 2.
Step-by-step explanation:
The given functions are
[tex]f(x)=2ln(x)[/tex]
[tex]g(x)=4ln(x)[/tex]
[tex]g(x)=kf(x+b)+c[/tex]
Where, k is stretch factor, b is horizontal shift and c is vertical shift.
If k>1, then it represents vertical stretch and if k<1, then it represents vertical compression.
If b>0, then the graph of f(x) shifts b units left and if b<0, then the graph of f(x) shifts b units right.
If c>0, then the graph of f(x) shifts c units up and if c<0, then the graph of f(x) shifts c units down.
[tex]g(x)=2(2ln(x))[/tex]
[tex]g(x)=2f(x)[/tex]
The stretch factor is 2>0, it means the graph of f(x) vertically stretch by factor 2.
