Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Graph of the tangent function Y equal tangent X. By definition tangent is sine divided by cosine. So everywhere that cosine equals 0, it's going to be undefined because we can't divide by zero. So cosine X can never be 0. Thus tangents undefined anytime cosine is 0. If we drew in our cosine graph. What we're saying is there's an asymptote every time the graph of cosine crosses the X axis. We can never ever have 1 / 0 now because tangent is sine divided by cosine at the zero location. Sine is 0 when cosine is 1, so a point on our graph is going to be 0. If we think about, we need to get closer and closer and closer to the asymptotes and between 0 and π halves. All of my tangent values are positive, so I'm going to go from this intercept up and get closer and closer and closer to the asymptote zero to -π halves. All of my tangent values are negative, so I'm going to go through and I'm going to get closer and closer and closer to the asymptote tangents. Periodic. It means it repeats. So in between each of these asymptotes, we're going to do the same thing. And this is our graph of tangent. Obviously not to scale. Our domain here is going to be all reals except X can never equal π halves plus K π and our range is going to be negative Infinity to Infinity. Our period is π. Let's look at some examples of what happens if we shift things around. So Y equal a tangent Omega X plus Phi plus B. The A is a vertical stretch or shrink π divided by Omega is the new. In tangent, and in cotangent the periods are π. The phase shift is negative Phi over Omega. The vertical translation is B. So if we have Y equal tangent of one of the quantity one half X - π + 1. So this 3 is a stretch because it's a positive 3. The 1/2 X, well the original. Was π. Now we're going to divide it by 1/2. So our new. Is going to be 2π on a tangent. What that really means is that half of it's going to go to the one side of the X axis and half is going to go to the other side of the X axis because in tangent we always go through the origin. Now we also have a translation. So a phase shift if we think about one half X - π equaling 01 half X would equal π, X would equal 2π. So this is telling me that my graph of tangent which look like this is now going to shift to π. So this negative π, if I add 2π to it, is going to end up at π. This Π + 2π is now going to end up at 3 Pi. This origin which was 0 + 2π is now going to end up at 2π over here. So we've now done a phase shift, which is this blue piece. We need to stretch it by three. So instead of being one here π four tangent of π force is 1, which is a usual point of reference. Instead of this being a height of one, it's now been stretched to three. So in between 2π and three Pi would have been five Pi halves. If we put in five Pi halves here, 5 Pi halves times a half, five π fourths minus π, well π is 4 fourths. So five π -, 4 fourths π is going to give me one fourth Pi in here. So tangent and 1/4 Pi was the three. So 5 Pi halves is going to be 3 in between π and 2π or three Pi halves is going to be -3. Now the plus one is just going to take this entire graph and it's going to shift it up by 1. So now five Pi halves is going to have a point of four, the .2π is going to be 1, and the .3 Pi halves is going to be -2. The asymptotes shift up and down also, but asymptotes go on and on, so they're still at π and 2π. Graph of cotangent function. Y equal cotangent X which is cosine X divided by sine X. Thus sine X can never equal 0 because we can't divide by zero. So the cotangent X is undefined every time sine X is 0, and that occurs when X equal K π. Thus asymptotes occur in these locations. So if we were going to look at this graph, we would know every time sine was 0, we'd have an asymptote. So that occurs at 0 and that occurs at π and that occurs at 2π and negative π. It's periodic, so it's going to repeat every π. Now, between 0 and π halves, all six of the trig functions are positive. So we know that the Y values have to be positive. At Π halves, it's cosine divided by sine, and cosine is zero. At π halves, sine is one, 0 / 1 is 0. Between Π halves and π, the cotangent values are all negative. So our graph is going to look like this. It's going to repeat, and that's what our generic graph of cotangent X looks like. If we thought about it π force the cotangent of π force is 1, so we'd have π force, 1. The cotangent at three Pi force would be -1 so 3 Pi force, -1. And of course Pi halves is 0. We have our two asymptotes at 0 and π for this particular. The domain here is going to be all real values except X is never going to equal K Pi where K is an integer. The range is going to be all real numbers. This is also an odd function. If you look at the function, if you thought about picking it up and turning it around 180° or pivoting it around the origin, it would look exactly the same. Hence, it's an odd function. We have phase shifts, vertical translations, stretches, and shrinks. So A is going to be the vertical stretch or shrink. The new. Would be π divided by Omega. The phase shift is negative Phi over Omega, and the vertical translation is B. So let's look at an example. If we have Y equal -1 half cotangent three X - π + 2, the new. Is going to be π / 3 new. So if we thought about just doing this in steps, our first step would be from zero to π. Thirds would be our new. So halfway between would have to be our intercept, IE Pi 6. Halfway between that Pi twelfths would be our one for tangent of π twelfths here. And halfway between Pi 6 and π thirds would also be a tangent value of -1 without the stretches and whatnot yet. So halfway between here would be π three Pi twelves, 2π twelves, 3 Pi twelves, or π fours. OK, so our first rough sketch is like this, and all we've done is change the period. So far, our phase shift says that three X - π is going to equal 0, so 3 X equal π or X equal π thirds. So each and every one of those points is going to get moved over π thirds. So instead of having an asymptote at zero, we're going to have an asymptote at π thirds. Instead of having this π thirds asymptote, we're going to add π thirds and we're going to be at 2π thirds instead of Pi 6 Pi 6 + π thirds π force Pi twelfths plus π thirds. Remember, we have to get a common denominator. So Pi twelfths +4 Pi twelfths would give me 5 Pi twelfths here, and π force plus the Pi 12th would give me 3 Pi twelfths plus Pi twelfths, or four Pi twelfths. Let's look at an example. So Y equal -1 half cotangent three X -, 2π + 1. The first step is going to be to find the new. So Π / 3. The old period of cotangent is π. The coefficient here is 3. So Pi third is the new. So we're going to do this just in pieces if our new. Is Π thirds. Here is 0 and here is π thirds. If we wanted to put in all of our other pieces halfway between 0 and π thirds, is π six. Halfway between zero and Pi 6 would be π twelfths. Halfway between Pi 6 and π thirds would be π fours. Remember, we take Pi 6 and we add π thirds and we divide by 2. So Pi 6 + 2, Pi 6 / 2, three Pi 6 / 2. Well, three Pi 6 is really just Pi halves. If we divide by two, it's the same thing as multiplying by 1/2 or π force. So our first thing that we just did was we just got our new. So all we've done so far is the period. Next we're going to do the phase shift. Three X -, 2π = 0 three X equal to π X equal to π thirds. What this is going to do is it's going to take those five key points and it's going to shift them. So I'm going to take 0 and add 2π thirds. I'm going to take π twelfths and add 2π thirds. I'm going to take Pi 6 and add 2π thirds. I'm going to take π fours and add 2π thirds, and we're going to take π thirds and add 2π thirds. So we have 2π thirds here. If we have twelfths, we're going to get a common denominator. So Pi twelfths plus this one's going to turn into 8 Pi twelves or 9 Pi twelves, which is 3 Pi fours. Here Pi 6 + 4 Pi 6 is five Pi 6. Here, 3 Pi twelves plus 8 Pi twelves is going to give us 11 Pi twelves. So our phase shift is now going to have an asymptote at 2π thirds. One of our key points is going to be at three π fourths. And right now, before we do our vertical stretch or shrink, that three π fourths would just be 1/5 Pi 6 is going to be our zero 11 Pi twelfths without the vertical stretch or shrink is going to be -1 and π thirds plus 2π thirds or π is going to be our other asymptote. So piece by piece, that's my phase shift. Now the vertical stretch or shrink a negative means it's going to be reflected around the X axis, the 1/2. So instead of being out at 1 back here, we're going to now be at 1/2. So 2π thirds out to π we're going to reflect it and we're going to shrink it. So at this location, my three Pi fours is now really going to have a value out of negative 1/2. My 5 Pi 6 is still 0. My 11 Pi twelfths is now going to come out at positive 1/2. Oh wow. Our final step is to add the one. So if we add the one, it really is just going to do a shift everything up by 1. So my key points are going to be 3 Pi fours 1/2 five Pi 6 one and 11 Pi 12th, 3 halves. Thank you and have a wonderful day.