Question

I need help with this practice I am having trouble with it The subject is trig

Answer 1

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given range

`(-\infty,-9\rbrack\cup\lbrack5,\infty)`

STEP 2: Find the cosecant function

`\begin{gathered} \text{The range of a cosecant function normally excludes the interval }(-1,1)\text{.} \\ The\text{ range in the question excludes the interval }(-9,5)\text{, which has a width 7 times as great.} \\ \text{Thus, we know the vertical factor is 7.} \\ \\ T\text{he midpoint of the excluded interval of the given function is }((-9+5))/(2)=-(4)/(2)=-2 \\ so\text{ that is the vertical translation.} \\ \text{The cosecant function normally has vertical asymptotes at }x=0\text{ and }x=\pi\text{ so the function is } \\ \text{expanded horizontally by a factor of }2. \end{gathered}`

Hence, the cosecant function is