what are the asymptotes for the graph of $y = \\sec(2x)-9$? select all that apply.

Explanation:

Step1: Rewrite the secant function in terms of cosine.

The function $y = \sec(2x) - 9$ can be written as $y = \frac{1}{\cos(2x)} - 9$.

Step2: Identify the condition for vertical asymptotes.

Vertical asymptotes occur where the denominator of the function is zero, provided the numerator is non-zero. For $y = \frac{1}{\cos(2x)} - 9$, the vertical asymptotes occur when $\cos(2x) = 0$.

Step3: Solve for the values of $2x$ where cosine is zero.

The cosine function is zero at odd multiples of $\frac{\pi}{2}$. $2x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

Step4: Solve for $x$.

Divide both sides by 2 to find the values of $x$. $x = \frac{\pi}{4} + \frac{n\pi}{2}$

Answer:

$x = \frac{\pi}{4} + \frac{n\pi}{2}$, where $n$ is an integer.