In Exercises 45–52, graph two periods of each function. y = se...

Question

In Exercises 45–52, graph two periods of each function. y = sec(2x + π/2) − 1

Accepted Answer

Welcome back. Everyone. In this problem, we want to sketch the graph of the function Y equals the second count of four X plus 1/4 of pi minus one using two period. Now, what do we already know? Well, that the second function is the reciprocal of the cosine. So if I had the graph of the cosine which looks something like this, then the graph of my second looks something like this in a sense it's reciprocated. Now, you might be saying, well, that's not exactly reciprocated what you're drawing right there in the blue. But that's because the C can function only works when the cosine of X is not equal to zero. And that makes sense because if the cosine of X equals zero, then the would be one divided by zero, which would have been undefined. So at those points instead, we have vertical asymptotes to represent where the cosine of X would have been zero. So that means if we want to draw the graph of the C can, we can go ahead and draw the graph of the cosine figure out where those asymptotes are and then remove and plot the graph of our C. So let's see if we can figure that out. So let's first start by letting Y equal the cosine of our value for X plus 1/4 of pi minus one. Now, what do we know about trigonometric graphs? Well, again, recall that for a trigonometric graph that's written in the general form a multiplied by the trick function. In this case, the cosine of BX minus C plus D where A would be the coefficient of the cosine, which here is one, B would be the coefficient of X which is four C would be the opposite number to the value inside the parentheses for the cosine. So that would be negative or fourth of pi and D will be equal to negative one. Then the amplitude of our graph that is how tall or smaller graphic would be equal to A which is one, the period for our graph would be equal to two pi divided by B. So in this case would be two pi divided by four. And that's a half of pi, our phase shift equals C divided by B which in this case would be negative or fourth of pi divided by four, which would be a negative or fourth of pi multiplied by a quarter which is negative 16th of pi. And our vertical shift would be equal to D which in this case is negative one. So right away, we have that information for our graph. We know that our graph isn't going to get any taller. We know that its period is going to be a half of pi, its phase shift is going to be a negative 16th of pi. In other words, we're going to shift it 1/16 of pi units to the left and its vertical shift is negative one. In other words, we're going to shift it down by one unit. So I'm going to go ahead and draw the cosine of my graph first for the period a half of pi and then we're going to shift our graph to the left and then below to see what it will look like. Now, you may have to revisit a previous lesson for curve sketching. But this is what our graph of the cosine of the, the cosine of the value would look like for a phase of a half of pi. So we've done that. We've, we've, we've shown our graph, we've shown the period. No, let's do the phase shift, let's shift it to the left by 1/16 of pi units. And then we're going to shift it down by one unit. So I'm gonna take my curve here and I'm going to shift it to the left by 1/16 of pi. Now, before I shift it down, it's really important for us to get those points where the would be and those points would be wherever the core sign of our value, wherever all of this would be equal to zero since we haven't shifted it yet. That's what our graph represents and it equals zero at the following points. Notice it equals zero. Here we are at a point pi 1/16 of pi. So let me draw my vertical line at that point. It also equals zero at 5 16 of pi which is right here. It does. So at 9 16 soft pie and I'm I I hope you're seeing those on the X axis as well. And it also does so at 13 16 of pi. So now we have all of those points where it is expected to be for our. So now we can go ahead and shift our graph down by one unit. And when we do that, notice that the cosine of our cosine ends up right here. No, I'm going to point out the some very critical points here which would be at our peaks and our valleys. OK? And these are the points that our are going to meet. So now if I were to draw the graph of the sea, it would look something like this. So remember it meets it the or it doesn't meet the. So here it would be a mirror fit like guys here. It's going to look something like this, then it would look like that between 9/16 of pie to 13/16 of pie. And then for a final leg from 13/16 of pie to 15th 16th of pie, it would look something like that where our points, our first point is 1/16 of pi and zero. Our next point is 3/16 of pie. I think I need a smaller, smaller pen. So 3/16 of pie and negative two at the 0.7 pi divided by 16, it would be zero at 11 16. So it would have been negative two and at 15/16 of pi it would have been zero. So this would be the sketch of our graph for Y equals the second of four X plus 1/4 of pi minus one. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 45–52, graph two periods of each function. y = sec(2x + π/2) − 1

Key Concepts

Period of a Trigonometric Function

Phase Shift

Graphing the Secant Function

Watch next