(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

How did they come up with [itex]\frac{1}{2}[/itex]+k for the equation of the vertical asymptote? I understand everything else except this.

3. The attempt at a solution

On this particular exercise, I graphed it and saw that each of my vertical dashed lines were all one whole unit apart. I've tried this method with another problem that had the dashed lines separated 2 units apart, took that 2, and multiplied it by the x=[itex]\prod[/itex]/2 + k[itex]\prod[/itex]. The result was [itex]\prod[/itex]+2k[itex]\prod[/itex], which was correct.

I tried it with other numbers and have gotten the correct answer, but I have a feeling I'm still doing something wrong. Because with this particular one using my method, I input [itex]\prod[/itex]/2 + k[itex]\prod[/itex] as my answer for the vertical asymptote which was incorrect. What am I doing wrong?

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# Trig: Writing the equation for vertical asymptote of a secant function?

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