Q. 13 - Precalculus Enhanced with Graphing Utilities

Question

For what numbers $x, -2\mathrm{\pi }\le \mathrm{x}\le 2\mathrm{\pi },$ does the graph of $y=secx$ have vertical asymptotes?

Step-by-Step Solution

Answer

The graph of $y=sec x$ have vertical asymptotes at  $-\frac{3\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2} \mathrm{in} \mathrm{the} \mathrm{interval} -2\mathrm{\pi }\le x\le 2\mathrm{\pi }.$

Step 1. Given Information

The given function is $y= sec x.$

We have to find the vertical asymptotes at $-2\mathrm{\pi }\le \mathrm{x}\le 2\mathrm{\pi }.$

Step 2. Finding the numbers

For the function, $y=sec x$ vertical asymptotes occur at $x=\frac{\mathrm{\pi }}{2}+\mathrm{n\pi }$.

So, for $-2\mathrm{\pi }\le \mathrm{x}\le 2\mathrm{\pi }$ the vertical asymptotes occur at $-\frac{3\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}.$