problems 5 - 16, sketch the graph of the function. include two full periods. 5. y = - sec x c/b : 2π/b :

Explanation:

Step1: Recall secant - function properties

The general form of a secant function is $y = A\sec(Bx - C)+D$. For the function $y=-\sec x$, we have $A=- 1$, $B = 1$, $C = 0$, and $D = 0$.

Step2: Find the period

The period of the secant function $y = A\sec(Bx - C)+D$ is given by $T=\frac{2\pi}{|B|}$. Since $B = 1$, the period $T=\frac{2\pi}{1}=2\pi$.

Step3: Find the vertical asymptotes

The secant function $y=\sec x=\frac{1}{\cos x}$. The vertical asymptotes of $y = \sec x$ occur where $\cos x=0$. For $y =-\sec x$, the vertical asymptotes are at $x=(2n + 1)\frac{\pi}{2}$, where $n\in\mathbb{Z}$. In the interval $[0,2\pi]$, the vertical asymptotes are at $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$.

Step4: Find key - points

We know that $\sec(0)=1$, so $y =-\sec(0)=-1$. Also, $\sec(\pi)=-1$, so $y =-\sec(\pi)=1$.

Step5: Sketch the graph

We start by drawing the vertical asymptotes at $x=\frac{\pi}{2},\frac{3\pi}{2}$ in the first period $[0,2\pi]$ and at $x=\frac{5\pi}{2},\frac{7\pi}{2}$ in the second period $[2\pi,4\pi]$. Then we plot the points $(0, - 1)$ and $(\pi,1)$. The graph of $y =-\sec x$ is a reflection of the graph of $y=\sec x$ about the $x$ - axis. The graph oscillates between $y =-\infty$ and $y=-1$ and between $y = 1$ and $y=\infty$ in each period.

For $\frac{C}{B}$, since $C = 0$ and $B = 1$, $\frac{C}{B}=0$. For $\frac{2\pi}{B}$, since $B = 1$, $\frac{2\pi}{B}=2\pi$.

Answer:

$\frac{C}{B}=0$; $\frac{2\pi}{B}=2\pi$; Graph: Sketch vertical asymptotes at $x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$, plot points $(0,-1),(\pi,1)$ etc. and draw the graph as a reflection of $y = \sec x$ about the $x$ - axis for two periods ($[0,2\pi]$ and $[2\pi,4\pi]$).