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

QUESTION POSTED AT 01/06/2020 - 04:31 PM

Answered by admin AT 01/06/2020 - 04:31 PM

\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
% 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}}

\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}
\textit{horizontal shift of }\cfrac{C}{B}\implies \cfrac{-1}{1}\implies -1
\textit{A is negative, so is flipped upside-down}
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