Respuesta :
[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}\qquad [/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \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{function period or frequency}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}[/tex]
now with that template in mind
[tex]\bf y=5+3sin\left(2x-\frac{\pi }{6} \right)\iff \begin{array}{lllccll} y=&3sin(&2x&-\frac{\pi }{6} )&+5\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array} \\\\\\ \textit{horizontal/phase shift by }\cfrac{C}{B}\implies \cfrac{-\frac{\pi }{6}}{2}\implies -\cfrac{\pi }{12}\impliedby \textit{to the right}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \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{function period or frequency}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}[/tex]
now with that template in mind
[tex]\bf y=5+3sin\left(2x-\frac{\pi }{6} \right)\iff \begin{array}{lllccll} y=&3sin(&2x&-\frac{\pi }{6} )&+5\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array} \\\\\\ \textit{horizontal/phase shift by }\cfrac{C}{B}\implies \cfrac{-\frac{\pi }{6}}{2}\implies -\cfrac{\pi }{12}\impliedby \textit{to the right}[/tex]
Before finding the phase shift we must simplify the equation.
y = 5 + 3Sin(2x - pi/6)
The x Inside the parenthesis has to be 1 x so we have to factor out a 2 from the parenthesis.
y = 5 + 3Sin2(x - pi/12)
The phase shift is pi/12 to the right
y = 5 + 3Sin(2x - pi/6)
The x Inside the parenthesis has to be 1 x so we have to factor out a 2 from the parenthesis.
y = 5 + 3Sin2(x - pi/12)
The phase shift is pi/12 to the right