OK, we want to graph this function. So the easiest way that I can tell you is to graph each piece separately and then look at the domain restrictions to come up with the entire graph. So when we want to do this seeking graph, remember that's just one over cosine. So our cosine graph looks something like this. Well, we can never have 1 / 0, so every time the cosine graph crosses the X axis, we're going to have an asymptote because that's where it would be undefined. And the secant graph's going to share the highest and lowest point because 1 / 1 is one and 1 / -1. So this is going to be our secant graph here. Now, zero for our Y value at Pi halves is really just a point. It's the point when X is π halves, Y is 0 tangent X. When we have our tangent graph, we know that our tangent is really sine divided by cosine. So that denominator can never be zero. So it's zero at π halves, three Pi halves, negative π halves, 5 Pi halves, etcetera. So let's say that's negative π halves, and this is π halves here, and this is 3 Pi halves out here. Now all six of the trig functions are positive in the first quadrant. So between 0 and π halves, my tangent has to be positive. It's got to be close to the asymptote. And at zero, we know that 0 / 1 is 0. So it's going to actually go through the origin and between zero and negative π halves on our unit circle. We know our tangent's negative, so it's going to come down like so tangent repeats every π. So we're going to have this repeating. So now when I put the three pieces together, we went from zero to π halves for secant. So we want this portion here. Then we want just a single point at Pi halves, and then from Pi halves to π we want the tangent. So when we put these three pieces together, we have this piece, we have a solid point, and we have this piece. So our answer is going to be C.