Question

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

Short answer

The graph of $y=secx$have vertical asymptotes atrole="math" localid="1646401597898" $-\frac{3\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}\mathrm{in}\mathrm{the}\mathrm{interval}-2\mathrm{\pi }\le x\le 2\mathrm{\pi }.$

Step-by-step solution

Step 1. Given Information

The given function is $y=secx.$

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, $y=secx$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$-\frac{3\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2}.$