which of the following is the graph of $y = 0.5sec(x+\frac{pi}{3})-2$?
which of the following is the graph of $y = 0.5sec(x+\frac{pi}{3})-2$?
Answer
Explanation:
Step1: Recall secant - function properties
The general form of a secant - function is $y = A\sec(Bx - C)+D$. For the function $y = 0.5\sec(x+\frac{\pi}{3})-2$, we have $A = 0.5$, $B = 1$, $C=-\frac{\pi}{3}$, and $D=-2$.
Step2: Find the vertical shift
The value of $D=-2$ means the graph of the basic secant function $y = \sec(x)$ is shifted down 2 units.
Step3: Find the phase - shift
The phase - shift is given by $\frac{C}{B}$. Here, $\frac{C}{B}=-\frac{\pi}{3}$, so the graph is shifted to the left by $\frac{\pi}{3}$ units.
Step4: Analyze the amplitude
The amplitude of the secant function (which is not really an amplitude in the same sense as for sine and cosine, but a vertical stretch factor) is $|A| = 0.5$. This means the vertical distance between the minimum and maximum values of the function is affected. The basic secant function $y=\sec(x)$ has a range of $(-\infty,-1]\cup[1,\infty)$. For $y = 0.5\sec(x+\frac{\pi}{3})-2$, the range is $(-\infty,-2 - 0.5]\cup[-2 + 0.5,\infty)=(-\infty,-2.5]\cup[-1.5,\infty)$. We can also find the vertical asymptotes of the function. The secant function $y=\sec(x)=\frac{1}{\cos(x)}$ has vertical asymptotes where $\cos(x)=0$. For $y = 0.5\sec(x+\frac{\pi}{3})-2$, the vertical asymptotes occur when $\cos(x+\frac{\pi}{3}) = 0$. $x+\frac{\pi}{3}=(2n + 1)\frac{\pi}{2}$, where $n\in\mathbb{Z}$. Solving for $x$ gives $x=(2n + 1)\frac{\pi}{2}-\frac{\pi}{3}=\frac{3(2n + 1)\pi-2\pi}{6}=\frac{(6n + 3 - 2)\pi}{6}=\frac{(6n+1)\pi}{6}$, $n\in\mathbb{Z}$. When $n = 0$, $x=\frac{\pi}{6}$; when $n=-1$, $x=-\frac{5\pi}{6}$.
We can then check the given graphs against these properties (vertical - shift, phase - shift, range, and vertical asymptotes) to determine the correct graph.
Since no other graphs are provided, we assume the graph that matches the above - derived properties is the correct one.