match the trigonometric function with one of the graphs i - vi. f(x)=2 sec(x)

Explanation:

Step1: Recall secant - cosine relationship

We know that $\sec(x)=\frac{1}{\cos(x)}$, so $y = 2\sec(x)=\frac{2}{\cos(x)}$. The vertical asymptotes of $y = \sec(x)$ occur where $\cos(x)=0$. For $y = 2\sec(x)$, the vertical asymptotes are at $x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$.

Step2: Analyze the amplitude - like property

The factor of 2 in $y = 2\sec(x)$ means that the minimum and maximum values of the function (in the non - asymptote regions) are $y = 2$ and $y=- 2$ (since the range of $\sec(x)$ is $(-\infty,-1]\cup[1,\infty)$ and for $y = 2\sec(x)$ it is $(-\infty,-2]\cup[2,\infty)$).

Step3: Evaluate at key points

When $x = 0$, $\cos(0)=1$, so $y = 2\sec(0)=2$. When $x=\pi$, $\cos(\pi)=-1$, so $y = 2\sec(\pi)=-2$.

Answer:

III