4. Graphing Trigonometric Functions

Graphs of Tangent and Cotangent Functions

Trigonometry

4. Graphing Trigonometric Functions

Graphs of Tangent and Cotangent Functions

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Hey everyone. So here we have an example where we are asked to graph the function $y = \frac{1}{2} \times \tan (x - \frac{π}{2})$ . Now, to solve this problem, what I'm going to do is first graph the function $y = \frac{1}{2} \times \tan (x)$ . I'm going to ignore this $\frac{π}{2}$ for now, but we will get to this later. So let's start with this function. To find $y = \frac{1}{2} \tan (x)$ , let's just recall what the tangent graph looks like. I know for the tangent, we have asymptotes at $- \frac{π}{2}$ , and at $\frac{π}{2}$ . Now, we do have more asymptotes on this graph; we're going to have other asymptotes at odd multiples of $\frac{π}{2}$ , but for now, I'm just going to draw these two. Now for the curve, I know the curve starts here at the middle and then goes up to the right and then down to the left. Another thing that I recall about the tangent is that we have points at $(\frac{π}{4}, 1)$ and $(- \frac{π}{4}, - 1)$ . The graph ends up looking something like this. Notice that we have a $\frac{1}{2}$ in front of our tangent. This $\frac{1}{2}$ , as you may recall, changes the amplitude or tallness of our graph. So rather than having these points at 1 and -1 on the y-axis, these are going to be reduced. So we're then going to have a point instead at $\frac{1}{2}$ and $- \frac{1}{2}$ for these two values of $\frac{π}{4}$ and $- \frac{π}{4}$ respectively. So that means the graph is going to look a little bit more like this, where we can see that we're a little bit shorter than we were before. So this right here would be the curve for $y = \frac{1}{2} \tan (x)$ .

Now, our last step for solving this problem is going to be to incorporate this $\frac{π}{2}$ . We need to figure out what $y = \frac{1}{2} \times \tan (x - \frac{π}{2})$ looks like. Well, something that I can see is that we have a $\frac{π}{2}$ here which is going to shift our graph in some sort of way. To find the shift, recall that we can find the $\frac{h}{b}$ ratio, which tells us how much we've shifted to the left or to the right. Now, see that we have a minus sign here, which means that this is actually a positive $h$ value. So we have $\frac{h}{b}$ is equal to just $\frac{π}{2}$ , because this one in the denominator is just going to keep this the same. So that means that our graph is going to be shifted $\frac{π}{2}$ units to the right because we got a positive $\frac{h}{b}$ . So what I can do is take this point right here and shift it $\frac{π}{2}$ units to the right. In fact, I can take this asymptote and shift it $\frac{π}{2}$ units to the right, which would put us at $π$ , then I can take this asymptote and shift it $\frac{π}{2}$ units to the right, which would put us at 0. So this is what the graph is going to look like. Because we have asymptotes at $π$ and then at 0, we're going to have another asymptote at $- π$ on the x-axis. What I can do from here is continue drawing these curves since they're going to repeat every $π$ amount of units. So we'd have another curve right back here, looking like this, and then like that. It's going to be the same as the curve that we had before because this curve just continues repeating. This is what our curves are going to look like. This is the graph for the tangent, and that is the solution to this problem. So, I hope you found this video helpful. Thanks for watching.

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