Find the Asymptotes y=-sec(7x)

For any ${\displaystyle y=\sec \left(x\right)}$, vertical asymptotes occur at ${\displaystyle x=\frac{\pi }{2}+n\pi }$, where ${\displaystyle n}$ is an integer. Use the basic period for ${\displaystyle y=sec\left(x\right)}$, ${\displaystyle (-\frac{\pi }{2},\frac{3\pi }{2})}$, to find the vertical asymptotes for ${\displaystyle y=-\sec \left(7x\right)}$. Set the inside of the secant function, ${\displaystyle bx+c}$, for ${\displaystyle y=a\sec (bx+c)+d}$ equal to ${\displaystyle -\frac{\pi }{2}}$ to find where the vertical asymptote occurs for ${\displaystyle y=-\sec \left(7x\right)}$.

Set the inside of the secant function ${\displaystyle 7x}$ equal to ${\displaystyle \frac{3\pi }{2}}$.

The basic period for ${\displaystyle y=-\sec \left(7x\right)}$ will occur at ${\displaystyle (-\frac{\pi }{14},\frac{3\pi }{14})}$, where ${\displaystyle -\frac{\pi }{14}}$ and ${\displaystyle \frac{3\pi }{14}}$ are vertical asymptotes.

The absolute value is the distance between a number and zero. The distance between ${\displaystyle 0}$ and ${\displaystyle 7}$ is ${\displaystyle 7}$.

The vertical asymptotes for ${\displaystyle y=-\sec \left(7x\right)}$ occur at ${\displaystyle -\frac{\pi }{14}}$, ${\displaystyle \frac{3\pi }{14}}$, and every ${\displaystyle x=\frac{\pi n}{7}}$, where ${\displaystyle n}$ is an integer. This is half of the period.

Secant only has vertical asymptotes.