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

AI Thread Summary
The discussion centers on understanding the vertical asymptote of the secant function, specifically for the equation y=4*sec(pi*x). The user graphed the function and noted that vertical asymptotes occur at intervals, leading to confusion about the correct formula. They discovered that the vertical asymptote can be derived by taking the reciprocal of the coefficient of x in the secant function and adjusting the standard formula accordingly. The relationship between the secant function and cosine is emphasized, as secant is the reciprocal of cosine, which aids in determining the asymptote's position. Ultimately, the correct approach involves recognizing the impact of the coefficient in the secant function's equation on the asymptote's calculation.
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Homework Statement


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Homework Equations


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

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=\prod/2 + k\prod. The result was \prod+2k\prod, 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 \prod/2 + k\prod as my answer for the vertical asymptote which was incorrect. What am I doing wrong?
 
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You would be right if the equation were y=4*sec(x). It's not. It's y=4*sec(pi*x). There's already a pi in the equation for y. y=4*sec(pi*x) doesn't have an asymptote at x=pi/2.
 
I noticed multiplying the vertical asymptote formula/equation by 1/pi cancels out the pi, resulting in that 1/2+k... but where did they get 1/pi from? Does that have any relation to secant being 1/cos?
 
Worked on another problem set up similarly and I think I got it!

I noticed that simply taking the 'B' (like in the y=Asin[B(x-C)]+D formula), turning it into the reciprocal (1/B), and thennnn multiplying it by \prod/2 + k\prod gets me the right asymptote. This *does* relate to inverse trig functions (ie, sec being the reciprocal of cos), right?
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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