Question

Find the phase shift of y=5+3sin(2x-pi/6)

Answer 1

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

Answer 2

`\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`





`\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}`


now with that template in mind



`\bf y=5+3sin\left(2x-(\pi )/(6) \right)\iff \begin{array}{lllccll} y=&3sin(&2x&-(\pi )/(6) )&+5\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array} \\\\\\ \textit{horizontal/phase shift by }\cfrac{C}{B}\implies \cfrac{-(\pi )/(6)}{2}\implies -\cfrac{\pi }{12}\impliedby \textit{to the right}`