Determine the simplest form of an equation for each graph. Cho...

Question

Determine the simplest form of an equation for each graph. Choose b > 0, and include no phase shifts. (Midpoints and quarter points are identified by dots.)

Accepted Answer

Welcome back. I am so glad you're here. We are asked to write an equation that represents the graph shown. The equation should be in simplest form. So B is greater than zero and the phase shift is not included. Then we're given a graph. I want to go over the answer choices first and then describe the graph in more detail. So our answer choices are answer choice A Y equals negative three plus four cotangent X answer choice by equals three plus cotangent of 1/4 X answer choice cy equals negative three plus four tangent of X. And answer choice dy equals three plus tangent of 1/4 X. Let's take a look at the graph. We have a vertical Y axis. A horizontal X axis. They come together at the origin in the middle. The range for what's drawn for our Y axis is from negative 10 to positive 10 in the domain. For what's drawn for hour, X axis is from negative three pi to positive three pi we have one period of a function shown here. We've got one line being drawn and there are vertical as some totes at X equals negative two pi and X equals a positive two pi. Then the function line is shown in between there because the distance from negative two pi to positive two pi is four pi our period is equal to four pi. And then the points which it passes through immediately to the right of that vertical Asymptote X equals negative to pi our function line is down toward negative infinity for the Y values. And then it's heading up along that vertical Asymptote. Then it curves over passes through the first identified point at negative pi two, then it passes through the 0.03 and then it passes through the point pie four. And then from pi four, it continues heading up toward positive infinity for the Y values along the vertical Asymptote X equals to pi. So there's our graph. Now we need to turn that into an equation. Well, when we've got that kind of stretched out s shape, we know that we're either dealing with a tangent function or a cotangent function and recall that the instructions told us there is no phase shift. So to identify if it's tangent or cotangent, we want to look at the Y axis, the tangent function is going to pass through the Y axis. If it's a normal tangent function, it's going to pass through at the origin at 00. For the cotangent function, it's going to have a vertical Asymptote at the Y axis where X equals zero. Well, this function line neither passes through the origin nor does it have a vertical Asymptote at X equals zero. So which one do we go with? Well, it does pass through the Y axis, it's just been shifted up. So we are dealing with a tangent function, but we can tell that there is some vertical shifting going on. We'll identify exactly how much in a little bit. So the standard equation for a tangent function is given by Y equals C plus A multiplied by the tangent of B multiplied by the quantity of X minus D. So essentially what we need to do is identify our ABC and D plug those in and then we'll have our equation. Now, we just need to compare our given function with the standard tangent of X function to see what's different, what has shifted, what has changed. All right. So start with period, the period of a standard tangent of X function is going to be just pie. And we find it with the equation that Pi divided by B and for a standard one B is equal to one. But our period as given is equal to four pi. So we can set four pi equal to pi divided by B. We can cross multiply to solve here. So that's going to be four pi B equals pi and then dividing both sides by pi we're going to have four B equals one, dividing both sides by four. We get a B value of 1/4. So our period for our um or excuse me, our B value is 1/4 and that something will plug in into our equation. Now, we just need to figure out AC and D. Now looking at the vertical asymptotes, the vertical asymptotes for a standard tangent of X function are going to be at X equals a negative pi divided by two and X equals a positive pi divided by two that does not match up with what we have here. But we know that that's because of our period, looking at the vertical asymptotes is important because it can indicate if there's been a phase shift left or right. But we were told in the instructions that there is no phase shift left or right. So that tells us our D value which indicates phase shift is equal to zero. Now, the last things here, the A and the C A is being multiplied by your function C is being added or subtracted to our function. So that's going to indicate a vertical change and we can figure out vertical changes by looking at our Y values. So let's compare the points of our given function with the points of a standard tangent of X function. Standard tangent of X function is going to pass through the points negative pi divided by four negative one, the 0.00 at the origin and the point pi divided by four positive one. So again, the X values are different because of our period being four pi but we can compare the Y values. So here the Y value for our standard, the first identified one is going to be at negative one and ours is up at two. So it's shifted up three units. Now for the next point, it goes from the origin and now it's all the way up at 03, it has again shifted up three units that's consistent. And the last one, again, the Y value would be at a positive one. But it's up here at a positive four, it has shifted up three units again. Now it is consistently shipping, shifting up three units. So what does that mean? It means for our A value what's being multiplied by our function? It has not been stretched and it has not been shrunk. And because it's heading in the same direction, it's heading up, it is not reflected over the X axis, essentially the A is not doing anything. So our, a value, what's being multiplied by the function is one, but it is up three consistently up three. So we have added three to our function and that's our C value because it's up three, that's a positive three. Our C value is equal to a positive three. And there we know our ABC and D and we can plug these in. So our equation is going to be Y equals C is three plus A is one, we don't have to write that tangent of B is 1/4 that's multiplied by X minus D. Well, D is zero minus zero is nothing. So it's just 1/4 X. Our answer is Y equals three plus the tangent of 1/4 X. We look at our answer choices and that matches with answer choice. D well done. We'll catch you on the next one.

Determine the simplest form of an equation for each graph. Choose b > 0, and include no phase shifts. (Midpoints and quarter points are identified by dots.)

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Key Concepts

Trigonometric Functions

Amplitude and Period

Graphing Trigonometric Functions

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