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 phase shift is not included. I'm going to describe the details on the graph more in a minute. But first, let's look at the answer. Choices, answer choice A is that Y equals the cotangent of five X. Answer choice by equals negative cotangent of X. Answer choice cy equals 1/5 the cotangent of X and answer choice dy equals five tangent X. All right, let's describe what's happening on the graph. We have a vertical Y axis, a horizontal X axis. They come together the origin, the range for what's drawn for our Y axis is from negative 10 to positive 10. And the domain for what's shown for our X axis starts at zero and goes to two pi divided by five. Now, along these graphs, we can see a few things. We've got two function lines repeating themselves. So we have two periods shown on our graph. One period for one function line goes from, we have first we have a vertical Asymptote, a first a vertical Asymptote at X equals zero. And then another vertical Asymptote at X equals pi divided by five, defining that first period. So the period length here from zero to pi divided by five is pi divided by five. Now there is another vertical Asymptote that helps define that second period at X equals two pi divided by five. And then what else is shown? So within one period, we'll look at the first one on here. First, we have kind of like stretched out s shaped curves and immediately to the right of that vertical Asymptote X equals zero. Our function line is heading up toward positive infinity along the Y values. And then it's coming down and it's curving and it first passes through the drawn point of pi divided by 20 one and then it passes through the point pi divided by 10 zero and then it passes through the point, the rep pi divided by 20 negative one. And then from that point, it curves and heads down toward negative infinity for the Y values along the vertical Asymptote X equals pi divided by five. Now, very similarly to the right of X equals pi divided by five. We've got our line heading up toward positive infinity for the Y values and then it's curving down, pass through the points. The first one here for this second function or the second line of our function is it pi divided by four one the second point drawn is at three pi divided by 10 0. And then the third one identified is at seven pi divided by 20 negative one. And then that line again continues down toward negative infinity for the Y values along the vertical Asymptote X equals two pi divided by five. All right, that's what we're looking at. So those long stretched out sort of s shapes are indicative of either being a tangent function or a cotangent function. And because we were told in the instructions that there is no phase shift, we can identify pretty quickly if it's going to be tangent or cotangent for tangent that would pass through the point 00, that's at the origin. And for cotangent that would have a vertical Asymptote at that Y axis at X equals zero. Here, we've got a vertical Asymptote at X equals zero. So we're talking about cotangent, the standard form of a line for cotangent would be Y equals C plus a multiplied by the cotangent of B multiplied by the quantity of X minus D. Now, essentially, we've just got to figure out our ABC and D plug those in and we'll have our equation for our line in order to identify ABC and D, we're going to compare what we know about our dr drawn function lines with what a standard Cotangent function line looks like. So we can start off with the period. The period is always found by taking pi and dividing that by the B value, a normal cotangent line would have a B value of one and its period would just be pi divided by one or pi. But we don't have a period of pi we have a period of pi divided by five. So our period is different from a normal cotangent function line. So what we can do is we set our period pi divided by five equal to pi divided by B and solve for B. Well, we can see here that B is equal to five. And so that's going to help change our function in our equation B is equal to five. OK. In a normal cotangent function, it would have a vertical Asymptote at X equals zero. And then it would have another vertical Asymptote one period later at PI and then another one at another period later to pi and so on here, we've got our vertical asymptotes at X equals zero. That's right. But then the next 11 period later is pi divided by five. So we already identified how that's changing our equation. That's for the B value. But what are the ver vertical asymptotes? Also tell us they can also tell us if there's a phase shift left or right, we already know from the instructions that we didn't have a phase shift. So our D value here is zero. Now let's figure out A and C what are the points that a cotangent function normally passes through and we'll just compare this to the first period of our function line. So normally it would pass through a point at pi divided by four one, then it would pass through the point pi divided by two zero and then it would pass through the point or excuse me, three pi divided by four negative one. The X values have been affected by the period. But the Y values are what we're looking at here to see if A or C is different because A is going to tell us if it's reflected over the X axis or if it's been stretched vertically and the C is going to tell us if there's a vertical shift up or down. Those are all Y value changes. And here, the Y values line up perfectly. It's still passing through a positive one with the first one. It's passing through a zero with the second one and it's passing through a negative one with the second one. So we haven't stretched it vertically, we haven't shrunk it vertically, we haven't moved it up or down. So because our um, a value tells us what it's being multiplied by and there's no change. Our A value is going to be one and our C value is what gets added on to say it's moving up or down and it's not moving up or down. Our C value is going to be zero. Now, let's fill these in and we'll have our equation Y equals C is zero. We don't need to write that down. A is one. We don't need to write that down because it's multiplied by. Then we have cotangent of B is five. And then for X minus DD is zero, we don't have to write that. So we can just have five X. And our equation of our line is Y equals the cotangent of five X. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

Struggling with Trigonometry?

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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Struggling with Trigonometry?

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.)<IMAGE>

Watch next

Graphs of Tangent and Cotangent Functions practice set

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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