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(a) Using pencil and paper, not a graphing utility, determine the amplitude, period, and (where appropriate) phase shift for each function.(b) Use a graphing utility to graph each function for two complete cycles. [In choosing an appropriate viewing rectangle, you will need to use the information obtained in part (a).](c) Use the graphing utility to estimate the coordinates of the highest and the lowest points on the graph.(d) Use the information obtained in part (a) to specify the exact values for the coordinates that you estimated in part (c).$$y=\sin (0.5 x-0.75)$$

the lowest points are : $\mathrm{C}(10.9,-1), \mathrm{D}(23.5,-1)$

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Heather Z.

Oregon State University

Alayna H.

McMaster University

Kayleah T.

Harvey Mudd College

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Video Transcript

So the equation we have is why is equal to sign off 0.5 x minus 0.75 This can be rewritten as why is he quick to sign? 0.5 times X minus 1.5. So from this equation, we can see that the aptitude is one. The period is two pi divided by 0.5 and that's he quickly to put foreplay and then the phase shift is 1.5 for part B. You can see the graph on the right side and for park si. You can see that the coordinates off the maximum point approximate coordinates are 4.64 to 1. And for the minimum point, the approximate corners A 10.9 to 5. Aunt Negative waas. So you can get this information also from the graph now for party. We want to find the exact partners off the maximum and minimum point. So we start with the graph for wise. He quit to sign off zero point fight X and what this graph looks like is this so here the period is four pi, which means that he made his group I and then the first peak is that pie in the first minimum is that a pie now for By is equal to sign 0.5 times X minus 1.5. What we would do is take this maximum point, shifted 1.5 units to the right, take this minimum point and shifted 1.5 minutes to the right. So the exact quarters off the maximum point would be pi class 1.5 and one and the exact orders off the minimum point would be a three pie. Last 1.5 and negative one.

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