Question

graph the given function over a one - period interval.
y = 1 - \frac{1}{4}\sec(x - \frac{3\pi}{2})
choose the correct graph below.
a.
b.
c.

Explanation:

Step1: Recall secant - function properties

The general form of a secant function is $y = A\sec(Bx - C)+D$. For the function $y = 1-\frac{1}{4}\sec(x-\frac{3\pi}{2})$, we have $A =-\frac{1}{4}$, $B = 1$, $C=\frac{3\pi}{2}$, $D = 1$. The period of the secant function $y = A\sec(Bx - C)+D$ is $T=\frac{2\pi}{|B|}$. Since $B = 1$, the period $T = 2\pi$.

Step2: Find the phase - shift

The phase - shift is given by $\frac{C}{B}$. Here, $\frac{C}{B}=\frac{3\pi}{2}$, which means the graph of $y =-\frac{1}{4}\sec(x)$ is shifted to the right by $\frac{3\pi}{2}$ units and then shifted up by 1 unit.

Step3: Analyze key points

The secant function $y=\sec x$ has vertical asymptotes at $x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$. For $y = 1-\frac{1}{4}\sec(x-\frac{3\pi}{2})$, the vertical asymptotes are $x-\frac{3\pi}{2}=(2n + 1)\frac{\pi}{2}$. Solving for $x$ gives $x=(2n + 1)\frac{\pi}{2}+\frac{3\pi}{2}=(n + 2)\pi$. When $x=\frac{3\pi}{2}$, $y = 1-\frac{1}{4}\sec(0)=1-\frac{1}{4}=\frac{3}{4}$.

Step4: Analyze the amplitude and reflection

The coefficient $A=-\frac{1}{4}$ means the amplitude of the secant - like curve is $\frac{1}{4}$ and there is a reflection about the $x$ - axis (due to the negative sign).

Answer:

Without seeing the actual graphs clearly, we can analyze the properties. The function $y = 1-\frac{1}{4}\sec(x-\frac{3\pi}{2})$ has a period of $2\pi$, a phase - shift of $\frac{3\pi}{2}$ to the right, a vertical shift of 1 unit up, an amplitude of $\frac{1}{4}$ and a reflection about the $x$ - axis. You need to check which graph has vertical asymptotes at appropriate positions, passes through the point $(\frac{3\pi}{2},\frac{3}{4})$ and has the correct shape considering the amplitude and reflection.