Question

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

Answer 1

$\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % function transformations for trigonometric functions % templates f(x)=Asin(Bx+C)+D \\\\ f(x)=Acos(Bx+C)+D\\\\ f(x)=Atan(Bx+C)+D \\\\ -------------------$

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

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

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

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.