In Exercises 53–60, use a vertical shift to graph one period of t...

Question

In Exercises 53–60, use a vertical shift to graph one period of the function.y = −3 sin 2πx + 2

Accepted Answer

Welcome back. I am so glad you're here. We're asked to sketch the graph of the following function. Consider only one period. Our function is Y equals negative six sign of open parentheses, four PX, closed parentheses minus five. Then we have a blank graph. We have a vertical Y axis and a horizontal X axis which come together at the origin. The domain for what's shown for our X axis is from negative 0.1 to 0.6. And the range for what's shown for our Y axis is from negative 12 to positive 12. All right. So we look at our function and we can see that this is in the format of Y equals a sign of open parentheses. BX minus C closed parentheses plus D and we can identify our A's and B's and C's and D's our A is what's being multiplied by our sign A here is negative six. Our B is what's being multiplied by the XB is four pi C is what's being added or subtracted directly from the X and there is nothing there. Our C term here is zero and D that's what's being added or subtracted after our sign part and here our D is equal to a negative five. And now to graph that information, our A SBS CS and Ds is very helpful. We recall from previous lessons that we're going to begin graphing by identifying, well, a bunch of things, we need to identify the amplitude. The amplitude is found by taking the absolute value of A A here is negative six. So we're taking the absolute value of negative six and that equals a positive six. All right, we've got the amplitude. Next, we want to identify the period. We recall from previous lessons that the period is going to be two pi divided by B keeping that two pi in the denominator, our B is four pi. So we have two pi divided by four pi which simplifies to one half. So our period is one half. Now for a phase shift, the phase shift would be given by C divided by B and we have no C terms. So that's going to be a zero in the numerator zero divided by anything. But here zero, divided by four pi is just going to be zero, there is no phase shift. And then finally, we do have a vertical shift because we have that D term, the vertical shift is found by taking our D. And because here our D is negative, it's going to be a vertical shift down. And then what's left is that five units? So we're going to shift down five units. All right. So we've got our amplitude or period, our phase shift and our vertical shift. What else do we need to know? Well, we know that when we have a vertical shift, our maximum is going to be given by D plus the absolute value of A. So here D is negative five plus the absolute value of A, we said was six, negative five plus six is a positive one. So our maximum is going to be at one for our minimum, that's going to be D minus the absolute value of A. And again, that's a negative five and now minus six, negative five minus six is a negative 11. Our minimum is going to be at negative 11. Other things that we can discern. All right, if we had a sine function that had no phase shift and no vertical shift, it would begin at 00 at the origin sine functions begin at the origin. But because we have that vertical shift down, we're going to be subtracting five from our Y value. And we're still going to start at zero for our X value, but zero minus five for the Y value is a negative five. So this will begin at zero, negative five. And the last thing that we can identify is how are X values for our five key points? Um How we're going to determine the change in X values which is through quarter periods. And so we take our period which is one half and we're going to divide that by four. Now, what's given to us on this graph is in decimals. So I'm going to write one half as 0.5. And then we're going to divide that by four, 0.5 divided by four is 0.125. So those are the quarter periods that we're going to use. So that was all step one. Now, let's work on step two, which is identifying our five key X values. That's going to be our starting point, our maximum and our minimum. So we'll have X sub one, X sub two, X sub three, X sub four and X sub five, we only need to consider one period. So we only need those five X values, as we said before, with this sine function, it's going to begin with an X value at zero and then we are going to add quarter periods and zero plus 0.125 is 0. 0.125 plus 0.125 is 0. 0.25 plus 0.125 is 0.375 and 0.375 plus 0.125 is 0.5. There are five key X values. Now, we want to find for step three, the associated Y values. So we have our Y sub one, Y sub two, Y, sub three, Y sub four and Y sub five. All right, we already said that this sine function with no phase shift but with a vertical shift is going to begin at zero negative five. So that first Y value is at negative five. Now, because our A value is negative, then our Y sub two is going to be a minimum. We're going to take our, our D value, which we've already done and we're subtracting the absolute value of our A. So we're subtracting our amp amplitude which gets us A Y sub two value of negative 11. Now, the thing is you can always take the X value. So here X sub two is 0.125 and you can plug it into our original function and you would get this solution of A Y sub two for negative 11. But because we understand what's going on with our sign function, we don't need to do that, but you can do that. All right. So we've gone down for a minimum. Now, we're going to get back up to our baseline for our Y sub three, which is at negative five. Next, we're going to have a maximum. We've already identified that that's a value of one. And then for Y sub five, we are back to negative five. So step four is just to graph, let's plot our points. The first one is at zero, negative five from the origin, we're going down five units. The next point is 0. negative 11 from the origin, we go to the right 0.125 and then we are going to go down 11 units. The next point is 0.25 negative five from the origin. We're going to the right 0.25 and down five. The next point is 0.3751 from the origin. We go to the right 0.375 and then we go up one, the last point is 0.5 negative five from the origin. We go to the right 0.5 and down five. And now we want to connect these points as smoothly as possible. And there we have our Y equals negative six sign of four PX minus five. Well done. We'll catch you on the next one.

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In Exercises 53–60, use a vertical shift to graph one period of the function.y = −3 sin 2πx + 2

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