which of the following is the graph of the function y = 2.5sec(x - 5)?

Explanation:

Step1: Recall secant - function properties

The general form of a secant function is $y = A\sec(Bx - C)+D$. For the function $y = 2.5\sec(x - 5)$, we have $A = 2.5$, $B = 1$, $C = 5$, and $D=0$. The amplitude of the secant function $y = A\sec(Bx - C)+D$ is not applicable in the traditional sense like for sine and cosine, but the vertical stretch factor is $|A|$. Here, $|A|=2.5$, which means the graph of $y=\sec(x)$ is vertically stretched by a factor of 2.5.

Step2: Consider the phase - shift

The phase - shift of the secant function $y = A\sec(Bx - C)+D$ is given by $\frac{C}{B}$. Since $B = 1$ and $C = 5$, the phase - shift is 5 units to the right.

Step3: Analyze key points

The secant function $y=\sec(x)=\frac{1}{\cos(x)}$ has vertical asymptotes where $\cos(x)=0$. For $y = 2.5\sec(x - 5)$, the vertical asymptotes occur when $x-5=(2n + 1)\frac{\pi}{2}$, $n\in\mathbb{Z}$, or $x=(2n + 1)\frac{\pi}{2}+5$, $n\in\mathbb{Z}$. When $\cos(x - 5)=1$, $y = 2.5$ and when $\cos(x - 5)=-1$, $y=-2.5$.

Answer:

Without seeing all the options, we cannot directly select the correct graph. But the correct graph should be a vertically - stretched (by a factor of 2.5) and right - shifted (by 5 units) version of the standard secant function $y = \sec(x)$ with vertical asymptotes at $x=(2n + 1)\frac{\pi}{2}+5$, $n\in\mathbb{Z}$, and values of $y = 2.5$ when $\cos(x - 5)=1$ and $y=-2.5$ when $\cos(x - 5)=-1$.