In Exercises 17–30, determine the amplitude, period, and phase...

Question

In Exercises 17–30, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = sin(x − π)

Accepted Answer

Hello, everyone. We are asked to identify the amplitude phase shift and period of the given sign function then sketch its graph. By considering only one period. The function we are given is Y equals sign of X minus five pi we are given a coordinate plane to graph. On the first thing to recall is that the general format for a sine function is Y equals a multiplied by the sign of B X minus C. Our function is Y equals the sign of X minus five pi. So to begin with, we will find the amplitude M your sine wave usually goes up to one on the Y axis and down to negative one amplitude will tell us if that's going to change. So our amplitude is the absolute value of A and here we don't have anything visibly we can see in front of the word sign. So this means this is the absolute value of one, which is one. So this means our Y values will be as high as one and as low as negative one. Next, we're looking at our phase shift base shift can be found by doing C divided by B. So C in our case, if we match up our formula to our actual function, it appears to be five pi and that'll be divided by B which the value in front of X is one. So our phase shift is five pi and then our period, the period is two pi divided by B. So this will be two pi divided by one. So our period is two pi. So next, let's start thinking about graphing this. I recall that I have four equal sections when I am graphing a sine wave. So I'm gonna take our period and divide it by four. So two pi divided by four, this means the width of each interval is gonna be pi divided by two. Also, we want to find our X one value first. So if you recall sine waves start at zero on the X axis, but we're gonna do zero plus our phase shift which is five pie. So our first X value is gonna be at five pi. So I'm gonna start my X Y table. So my first X value is five pi and since this is a normal sine wave, Y will be zero at that point for our next X value. Why will increase to the amplitude of one? So now let's find our Y two. So that is our Y one. So five pi plus the width just pi divided by two. And this looks like it'll be 11 pi divided by two. Next our wave will go back to zero on the Y. And to find the third X, we have 11 pi divided by two plus the width which was pi divided by two, which gives us 12 pi divided by two, which is equal to six pie. So I'm gonna put six pie in here for my next one. The Y value dips down to negative one and we can find this value by doing, I'm gonna treat it as 12 pi divided by two still and add pi divided by two. So that's 13 pi divided by two. And our last X value we will do 13 pi divided by two. And we're gonna add another pie divided by two, just 14 pi divided by two which is seven pi and that's when we will go back to our X axis and Y will be zero. So we have all our values. Let's put them on our graph. So at five pi Y is zero at 11 pie over two, Y is one. So we could see that's between five pie and six pie. So I'm gonna put that right here. At six P Y is zero at 13 pi divided by two, which I can see again is between six pi and seven pi. We're going down to Y equals negative one and at seven pi we're back up at Y equals zero. So connecting these with a wave, we have our sine wave over one period of two pi with a five pi phase shift and an amplitude of one have a nice day.

In Exercises 17–30, determine the amplitude, period, and phase shift of each function. Then graph one period of the function. y = sin(x − π)

Key Concepts

Amplitude of a Trigonometric Function

Period of a Sine Function

Phase Shift of a Trigonometric Function

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