1) Find the Asymptotes:
y = -2 sec 2pi x
pi/2b and 3pi/2b
pi/2(2pi) = pi/4 and 3pi/4
Is this correct?
2) Find the period and asymptotes:
y = 1/2 sec pi x / 2
b = pi/2 I do not know how to find the period of asymptotes for this one. My original answer was 1 for the period and pi and 3pi for the asymptotes.
No.
y = -2 sec 2pi x
when the argument is PI/2, or 3PI/2 of the secant function, it is inf
2PI x =PI/2
x= 1/4 or x= 3/4 is the x value of the asymptote.
2. PIx/2=PI/2 or 3PI/2
x=1 or x=3 is asympotes
period? here is a neat way to memorize finding frequency and period.
You should know this could also be written as sin w*x, where w(omega) is the angular frequency (2PI f) or 2PI/period
So where we have sec PIx/2
convert it PI*x/2=2PIx/2 * 1/period
period= 1/f= 1
To find the asymptotes of a function, you need to consider the values that make the function undefined. In the case of secant functions, the function is undefined when the cosine is equal to zero. The cosine function is zero at \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\), \(\frac{5\pi}{2}\), and so on.
For the first example, \(y = -2\sec(2\pi x)\), the period of the secant function is \(2\pi\). To find the asymptotes, you need to divide \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) by the coefficient of \(x\), which is \(2\pi\).
\(\frac{\pi}{2(2\pi)}\) simplifies to \(\frac{1}{4}\), and \(\frac{3\pi}{2(2\pi)}\) simplifies to \(\frac{3}{4}\).
So, the correct asymptotes for \(y = -2\sec(2\pi x)\) are \(x = \frac{1}{4}\) and \(x = \frac{3}{4}\).
For the second example, \(y = \frac{1}{2}\sec\left(\frac{\pi x}{2}\right)\), the coefficient of \(x\) is \(\frac{\pi}{2}\).
To find the period, you divide \(2\pi\) by the coefficient of \(x\), which gives you \(\frac{4\pi}{\pi}\), which simplifies to 4.
The asymptotes for secant functions occur when the cosine is zero, which happens at \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\), \(\frac{5\pi}{2}\), and so on.
However, in the given function \(y = \frac{1}{2}\sec\left(\frac{\pi x}{2}\right)\), the coefficient in front of \(x\) is \(\frac{\pi}{2}\), so we need to divide the values of the asymptotes by that coefficient.
\(\frac{\pi}{2}\) divided by \(\frac{\pi}{2}\) simplifies to 1. So, one of the asymptotes is \(x = 1\).
\(\frac{3\pi}{2}\) divided by \(\frac{\pi}{2}\) simplifies to 3. So, another asymptote is \(x = 3\).
In summary, for the function \(y = \frac{1}{2}\sec\left(\frac{\pi x}{2}\right)\), the period is 4 and the asymptotes are \(x = 1\) and \(x = 3\).