### Video Transcript

Which of the following is the graph of π¦ equals tan π₯?

Letβs begin by recalling some key properties of the tangent function. tan π₯ is periodic with a period of 180 degrees. So, the same pattern on its graph repeats at every interval of length 180 degrees. The graph of tan π₯ has vertical asymptotes, and these are positioned at 90 degrees plus any integer multiple of 180 degrees. These asymptotes occur when cos of π₯ is equal to zero. To help us with identifying which graph represents the tangent function, we can write out the position of some of these asymptotes explicitly. They occur at negative 90 degrees, 90 degrees, 270 degrees, and so on.

The final property that weβll recall is that the roots of the tangent function are the same as the roots of the sine function. And they are 180π degrees, where π is any integer. So, the roots occur at any integer multiple of 180 degrees. Letβs begin then by considering the first property, which tells us that the period of the tangent graph is 180 degrees. Looking at graph (A), we can see that this graph has a period of 90 degrees, and so we can rule this one out. Each of the other four graphs, though, do have the correct period of 180 degrees.

Letβs consider the second property, which tells us that we should see vertical asymptotes at 90 degrees plus 180π degrees. Now, as weβve already checked the periodicity of 180 degrees, it will be enough for us to check whether there is a vertical asymptote at 90 degrees. Options (C), (D), and (E) do all have a vertical asymptote when π₯ is equal to 90 degrees, but option (B) doesnβt. We can see that when π₯ is equal to 90 degrees, the value of the function in graph (B) is zero. So, for this reason, we can rule out option (B).

Next, we need to consider the roots of each graph. We know that the roots of the tangent function occur at 180π degrees, where π is any integer. Again, since weβve already checked the periodicity, it will be enough to check whether the graph has a root at π₯ equals zero. Options (D) and (E) do each have a root at π₯ equals zero. The graph crosses the π₯-axis at this value, but option (C) doesnβt. So we can rule out option (C).

Weβre left then with just two graphs, options (D) and (E), which each have the correct periodicity, the correct vertical asymptotes, and the correct roots. To distinguish between the two graphs, we note that the function in graph (D) is negative over the interval between zero and 90 degrees, while the function in graph (E) is positive in this interval. Angles between zero and 90 degrees are acute tangles or angles in the first quadrant of the unit circle. Since the tangent function is positive for acute angles, this means that graph (D) is not the graph of the tangent function.

We can remind ourselves why tangent is positive in the first quadrant if we recall that the coordinates of any point on the unit circle are cos π, sin π, where π is the counterclockwise angle between the positive π₯-axis and the radius connecting that point to the origin. In the first quadrant, both the π₯- and π¦-coordinates are positive. And so, cosine and sine are both positive. As the tan of an angle π is equal to the sin of π over the cos of π, it then follows that tangent is also positive in the first quadrant.

So, by recalling the key properties of the tangent function, we found that the graph of π¦ equals tan π₯ is graph (E).