graph the following function. show at least two cycles. use the graph to determine the domain and range of…

Explanation:

Step1: Recall the domain of (y = \sec x)

The function (y=\sec x=\frac{1}{\cos x}). The cosine function (\cos x = 0) when (x=\frac{\pi}{2}+k\pi=\frac{(2k + 1)\pi}{2}), where (k\in\mathbb{Z}). So the domain of (y = \sec x) is (\left{x|x\neq\frac{(2k + 1)\pi}{2},k\in\mathbb{Z}\right}). For the function (y=-9\sec x), the domain remains the same since the vertical - stretch and reflection do not affect the values of (x) for which the function is undefined.

Step2: Recall the range of (y=\sec x)

The range of (y = \sec x) is ((-\infty,- 1]\cup[1,\infty)). For the function (y=-9\sec x), we use the transformation (y = A\sec x) (here (A=-9)). When (y=\sec x), if (y_1=\sec x), then (y=-9y_1). If (y_1\geq1), then (y=-9y_1\leq - 9); if (y_1\leq - 1), then (y=-9y_1\geq9)

Answer:

The domain of (y = - 9\sec x) is (\left{x|x\neq\frac{k\pi}{2},k\text{ is an odd integer}\right}) (Option A). The range of (y=-9\sec x) is ((-\infty,-9]\cup[9,\infty))