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 going to be greater than zero and phase shift is not included. Then we've got a graph. I'm going to describe that in more detail in a moment. First, let's take a look at our answer. Choices. Answer choice A is that Y equals four tangent X answer choice by equals negative two cotangent X answer choice cy equals negative four tangent X and answer choice dy equals negative two tangent X. All right. Let's take a look at the graph. We have a vertical Y axis and horizontal X axis. They come together at the origin. The range for what's drawn for our Y axis goes from negative 10 to positive 10. And the domain for what's drawn for our X axis is from negative pi divided by two to positive three pi divided by two. Now let's describe what's going on with this function. So here we can see that the period for our function is going from negative pi divided by two to positive pi divided by two and then it repeats itself. There are two periods drawn here and the period for this function, if we take from negative pi divided by two to positive pi divided by two, that equals pi, we can see that we have vertical asymptotes at the X values of negative pi divided by two, X equals positive pi divided by two and X equals positive three pi divided by two. And then we have two kind of elongated, stretched out s figures inside of each of one inside of each of those two different periods between the three vertical asymptotes. So if we look at the first period from negative pi divided by two to pi divided by two, the, the function line is heading up toward positive infinity along the vertical Asymptote X equals negative pi divided by two to the right of that. And then it starts heading down and it passes through three points that are pointed out here. The first point it passes through that's drawn on here is negative pi divided by four, four. The next point passes through is the origin 00 and then it passes through the point positive or excuse me, yes, positive pi divided by four, negative four. And then for the rest of that line, that function line is heading down toward negative infinity. For the Y values along the vertical Asymptote X equals pi divided by two. And then that shape repeats itself for the second period to the right of the vertical Asymptote pi divided by two. The function line is heading up toward positive infinity for the Y values it comes down passes through three points. The first one is three pi divided by four, four. The next one it passes through is at pi zero. And then the final one it passes through is five pi divided by four, negative four. Then it heads down toward negative infinity for the Y values along that vertical Asymptote X equals three pi divided by two. So that's what we're given here. Now, these long stretched out s figures are indicative of either having a tangent function or a cotangent function. So which one do we have? Well, the fastest way to identify this because we know that there is no phase shift. We know that the tangent function typically passes through the origin at 00. Whereas the cotangent function has a vertical Asymptote at X equals zero. So here we can see that our function lines pass through the origin. So we're talking about a tangent function, the standard form for our tangent function is going to be Y equals C plus a multiplied by the tangent of B multiplied by the quantity of X minus D. And so now we just need to figure out our ABC and D plug that in to our equation and we'll have our given equation. So what do we know? And how is this different from a normal tangent function. Well, let's go through and identify the period, vertical asymptotes and points that a normal tangent function would pass through the period for a normal tangent function is pi so we have the normal period here. How does that affect us? Well, the period is given by that B value in our ABC and D that we're looking for. So the period is typically found to be pi divided by B and because here we know that it's equal to pi pi divided by our B term is going to be one. So B is one, we got one of them figured out. Now, the vertical asymptotes are helpful because that's related to any sort of a phase shift. Here, we already know there is no phase shift and the vertical asymptotes of a standard tangent function are going to be at X equals negative pi divided by two and X equals a positive pi divided by two. So that one's fine that is related to our D value, the devalue tells us about any sort of phase shift. So because our vertical asymptotes are normal and because there's no phase shift, the D value is going to be equal to zero in the equation that we're writing. All right. Now, the points that a normal tangent function would typically pass through the normal tangent function is typically going to pass through the points here. Let me move this second periods info down for a second. And we'll compare it to the first period for the function. So the normal points that would pass through would be negative pi divided by four that matches up. And then the Y value would be negative one. The next point it would pass through would be 00, that one matches up. And then another point that it would pass through would be positive pi divided by four, that one matches and then it would be a positive one. So here we can see some differences with our points. Instead of passing through at the negative one, for the negative pi divided by four, you can draw that where that would be on the graph. And instead of passing through it positive pi divided by four positive one, then we can draw a line curving and connecting through there. We can see that our function line has been reflected over the X axis. And that's important, they will reflect over the X axis when the a value that's being multiplied by our tangent function is negative. And more specifically, it's stretched it out too. Instead of passing through pi divided by 41, it's passing through pi divided by four, negative four. So it's being multiplied by a factor of four. Our a value here is negative four. Now, the last part that we need to figure out is that part CC would be shown by whether or not it's passing through the origin because we're still passing through the origin. It hasn't shifted up or down. There isn't a vertical transformation happening. So our C value is going to be zero as well. Now, we can fill these in and we'll have our equation. So that's going to be Y equals C is zero. So we don't have to put that A is negative four multiply by the tangent of the B value is one. So we don't need to put that and we have our X minus DD is nothing. We don't have to put anything for that. And our equation is Y equals negative four tangent X. We look at our answer choices and that matches with answer choice. C 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

Basic Form of Sine and Cosine Functions

Period and Frequency of Trigonometric Functions

Phase Shift and Midpoint Identification

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