Question

What kind of shift does the graph have? y = -sin (x-1) from y = sin(x)

Answer 1

$\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\ % function transformations for trigonometric functions \begin{array}{rllll} % left side templates f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}} \\\\ f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\ f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}} \end{array} \\\\ -------------------$

$\bf \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}$

$\bf \bullet \textit{vertical shift by }{{ D}}\\ \left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\ \left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ \left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ \left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)$

now, with that template in mind, let's see

$\bf \begin{array}{lllll} y=&-1sin(&1x&-1)&+0\\ &A&B&C&D \end{array} \\\\\\ \textit{horizontal shift of }\cfrac{C}{B}\implies \cfrac{-1}{1}\implies -1 \\\\\\ \textit{A is negative, so is flipped upside-down}$