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Vlada [557]
3 years ago
9

How does the graph of y = sec(x + 3) – 7 compare with the graph of y = sec(x)?

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
4 0
\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% function transformations for trigonometric functions
% templates
f(x)=Asin(Bx+C)+D
\\\\
f(x)=Acos(Bx+C)+D\\\\
f(x)=Atan(Bx+C)+D
\\\\
-------------------

\bf \bullet \textit{ stretches or shrinks}\\
~~~~~~\textit{horizontally by amplitude } A\cdot B\\\\
\bullet \textit{ flips it upside-down if }A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }B\textit{ is negative}

\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{C}{B}\\
~~~~~~if\ \frac{C}{B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{C}{B}\textit{ is positive, to the left}\\\\
\bullet \textit{vertical shift by }D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}

\bf \bullet \textit{function period or frequency}\\
~~~~~~\frac{2\pi }{B}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
~~~~~~\frac{\pi }{B}\ for\ tan(\theta),\ cot(\theta)

with that template in mind,

\bf y=sec(\stackrel{B}{1}x+\frac{C}{3})\stackrel{D}{-7}

the derived function has a horizontal shift of C/B or +3/1 or +3, namely 3 units to the left.

and has a vertical shift D = -7, of 7 units downwards.
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