Determine the vertical asymptotes of the graph of graph of y=secx/logx for x≤2pi
Vertical asymptotes occur when the denominaotrs of your function equal zero.
y = secx/logx is the same as
y = 1/(cosx logx)
so you would have those asymptotes when
cosx = 0 or logx = 0
x = pi/2 or 3pi/2 or x = 1
To determine the vertical asymptotes of the graph of y = sec(x)/log(x) for x ≤ 2π, we need to find the x-values where the function approaches infinity or negative infinity.
First, let's consider the denominator, which is log(x). The logarithm function is only defined for positive values, so we need to find the x-values that make log(x) equal to zero.
Setting log(x) = 0, we get x = 1. Therefore, x = 1 is a vertical asymptote.
Next, let's look at the numerator, which is sec(x). The secant function takes on the value of infinity or negative infinity when its cosine component becomes zero. The cosine function is zero at odd multiples of π/2, which means that sec(x) is undefined (infinity or negative infinity) at x = (2n + 1)π/2 for any integer n.
So, we have additional vertical asymptotes at x = π/2, 3π/2, 5π/2, and so on.
Therefore, the vertical asymptotes of the graph of y = sec(x)/log(x) for x ≤ 2π are x = 1, π/2, 3π/2, 5π/2, and so on.