Answer: 


Solution Method: 
First, we want to recall the parent function of the sine graph which starts at the origin, increases to 1, then decreases to 0, -1, then back to 0.

Next, we want to identify the transformations of the parent sine graph. We have that \(A=3\) and \(B=2,\) and we do not have a C or D. Without a phase or vertical shift, we know to center our graph with respect to the x and y axes. 

Since this is a sine graph and I have no horizontal or vertical shifts, the first point I can plot is the y-intercept at the origin, (0,0). 

Next, I need to find the period. I know that \(B = \frac{2\pi}{\text{period}}\), so period=\(\frac{2\pi}{B} = \frac{2\pi}{2}=\pi\). The period of \(\sin(x)\) is \(2\pi\) so that is the difference between this x-intercept and this x-intercept. So for \(f(x)=3sin(2x),\) there is an x-intercept a period, \(\pi\) away from the origin point. And then there is another x-intercept halfway between those two at \(\frac{\pi}{2}.\) 

Finally, to plot the max and mins, I turn to the amplitude. \(A=3,\) so the maximum point is located up 3 from the central axis, halfway between the x-intercepts. Then the minimum is down 3 from the central axis, halfway between these intercepts. 

And those 5 points are all we need to graph a period of this function so we just connect the dots. 

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Transformations of Trig Graphs Exercise 1

Answer:



Solution Method:The first thing we're going to do is think about what one period of cosine looks like. And actually in this case, because there's this negative at the front of my function,so I'm thinking about what one period of negative cosine looks like. So one period of cosine has a y-intercept at (0,1), coming down to \((\pi,-1),\) and coming back up again to \((2\pi,1).\) Then a negative will reflect this over the x-axis. So we will use this as a template for what our graph should look like.

The next thing we're going to do is establish the graph's axes. In cases where we have vertical or horizontal shifts, those shifts determine the axes of the graph. We have that \(C=-2,\) left two, and \(D=-1,\) down 1. So let's draw our coordinate plane here, and then the axes of this graph are located at \(x=-2\) and \(y=-1.\)

Now that we have the axes, we want to plot our first point, the y-intercept. We have that \(A=4.\) So up 4 and down 4 from our horizontal axis will be the maximum and minimum of this graph. So the range of this function is\( [-5,3]\). Then we look at that sketch of negative cosine and see the y-intercept is an amplitude below the horizontal axis, so the first point we will plot is \((-2,-5).\) 

Now let's determine our period. If \(B=\frac{2\pi}{\text{period}}\), then \(\text{period}=\frac{2\pi}{B}=\frac{2\pi}{\frac{\pi}{3}}=6\). So one period is 6 units long. Soone period is 6, so from min to max is half that, 3, and the x-intercept between the min and max is 1.5 from the min. So now we can plot the rest of our points.

So 1.5 spaces right from \((-2,-5)\) is the intersection with the horizontal axis \((-.5,-1).\) Then another 1.5 spaces is our maximum \((1,3).\) Then another 1.5 is \((2.5,-1),\) and finally another min at \((4,-5).\) 

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Transformations of Trig Graphs Exercise 2

Answer: \begin{align*}
    f(x) &= -\sin(3x)+4 \\
    f(x) &= \cos\left(3\left(x+\frac{\pi}{6}\right)\right)+4
\end{align*} 

Solution Method: 


Sine and cosine graphs only differ by a phase shift, so any sinusoidal graph can be written as a sine and a cosine function. Furthermore, amplitude, period, and vertical shift will be the same for both functions. The only difference will be phase shift and possibly a reflection. 

The first thing we need to do is find the center axis. We can see that on this graph the center axis is \(y=4,\) because the maximums and minimums are equidistant from it. The location of this center axis tells us our vertical shift. Our axis is up 4 from the x-axis so \(D=4.\) We can also use this center axis to tell us amplitude. The maximum and minimums are up one and down 1 from the center axis, so our amplitude \(A=1.\) 

Next, we need to find the phase shift, C, for both sine and cosine.  We know that cosine intercepts the y-axis at a maximum. So I want to find the nearest maximum to the y-axis and that will tell me how far cosine is horizontally shifted. There is a maximum left \(\frac{\pi}{6}\) from the y-axis. So my phase shift for cosine is \(C=+\frac{\pi}{6}\).

 We know that sine intercepts the y-axis at the origin and then increases. Luckily for us, this graph intersects the y-axis at the center axis, but it decreases instead of increases. So I have two options: I can find the nearest intersection of the center axis that increases to the right, or I can reflect my sine graph and make the function negative, so it decreases first. I'm going to do the latter, so then my sine function is negative, and \(C=0.\)

Lastly, we need to find the period and calculate B. A period of sine starts at the center axis and has a minimum and maximum, so if I start here at the y-intercept, a minimum and maximum away puts the end of my cycle at \(\frac{2\pi}{3}.\) So the period of this graph is \(\frac{2\pi}{3}.\) Then to find B, we have \(B = \frac{2\pi}{\frac{2\pi}{3}} =3.\) 

So now we put it all together and get the equations
\begin{align*}
    f(x) &= -\sin(3x)+4 \\
    f(x) &= \cos\left(3\left(x+\frac{\pi}{6}\right)\right)+4
\end{align*}

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Transformations of Trig Graphs Exercise 3