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 }(C)/(B)\\ ~~~~~~if\ (C)/(B)\textit{ is negative, to the right}\\\\ ~~~~~~if\ (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}\\ ~~~~~~(2\pi )/(B)\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ ~~~~~~(\pi )/(B)\ for\ tan(\theta),\ cot(\theta)`

with that template in mind,


`\bf y=sec(\stackrel{B}{1}x+(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.

Answer 2

C

Explanation: