Sketch the two cycles of the graph (starting from x = 0) of the given function. Indicate the amplitude, period, phase shift, domain and range and vertical asymptotes.

Question

Sketch the two cycles of the graph (starting from x = 0) of the given function. Indicate the amplitude, period, phase shift, domain and range and vertical asymptotes.

y = -sec x

Answer 1

We can begin by defining some of the terms in the problem. The amplitude of a periodic wave function is half the distance between the highest point and the lowest point. The period of a repeating function is the horizontal distance from the beginning of one repeat to the beginning of another repeat. For a simple sinusoidal function like cos(x), the period is the distance from peak to peak or trough to trough. The phase shift is how horizontally displaced the wave is from normal. The domain is all possible x values for which the function is valid. The range is all possible y values for which the function is valid.

In this example, we must recognize that - sec(x) is - 1/cos(x). Let us consider sec(x) - also 1/cos(x) - first to make things simpler. We know that the function cos(x) is sinusoidal wave passing through the point (0,1) with an amplitude of 1, a period of 2π, a phase shift of 0, a domain of - ∞ to +∞ and a range of -1 to 1. If for example, we were presented with the function cos(x-2), then everything would be the same except the the graph would pass through point (2,1) so the phase shift would be 2 as the graph is shifted 2 units to the right.

If we know the characteristics of cos(x), then we know the characteristics of sec(x). Let us first consider the domain. Whenever we are dealing with dividing a function, which we are here since sec(x) is 1/cos(x), we have to watch out for when the denominator is 0 as 1/0 is undefined. We know that cos(x) is 0 at π/2 ±πk where k is any positive integer. Thus the domain of sec x is everywhere except at π/2±πk. Everywhere else, cos (x) is non-zero so sec (x) is defined. The vertical asymptotes of a function occur at x values where a function is undefined so these will also be at π/2±πk.

The range of sec(x) can also be derived from the range of cos(x). The maximum value of cos(x) is 1, the minimum value is -1 and the function approaches 0 and hits 0. We can think of the inverse step by step. As cos(x) goes from 1 to 0, 1/cos(x) goes from 1 to ∞. As cos(x) crosses 0, 1/cos(x) is undefined. As cos(x) goes below 0, suddenly 1/cos(x) becomes a very large negative number. As cos(x) then goes to -1, 1/cos(x) goes from -∞ to -1. As cos(x) goes from -1 back to 0, 1/cos(x) goes from -1 to -∞. Finally as cos(x) goes from 0 to 1, 1/cos(x) goes from ∞ to 1. I encourage you to graph cos(x) and sec(x) side by side to see the comparison visually. Looking back the values we discussed, the range of cos(x) is (-∞, -1] and [1, ∞). Note, parentheses indicated non-inclusive - ∞ is always noted as non-inclusive - while square brackets are inclusive - sec(x) actually hits 1 and -1.

Graphing sec(x), we see that the graph does not have an amplitude as it goes to ±∞. We also see that the period of sec(x) is the same as the period of cos(x), 2π. Like discussed earlier, the phase shift is 0 because the graph has not been shifted to the right or left. For example, sec(x+π/6) would have a phase shift of -π/6 as the graph is shifted to the left by π/6 units. I encourage you to test a few values out and graph the shifted function if this is unclear. As your function is -sec(x), all you need to do is flip the graph of sec(x) over the x axis.