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Virtual Math Learning Center Texas A&M University Virtual Math Learning Center
tags:
Amplitude Cosine Period Phase Shift Trigonometry Trigonometry Series

Transformations of Trig Graphs Exercise 2

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Graph a period of  \(f(x)=-4\cos\left(\dfrac{\pi}{3}(x+2)\right)-1.\)

Answer:
Graph of one period of a transformed sine function with a minimum value of -4 at -1 and 4, a max value of 3 at 1, and middle values of -1 at -0.5 and 1.5

Graph of one period of a transformed sine function with a minimum value of -4 at -1 and 4, a max value of 3 at 1, and middle values of -1 at -0.5 and 1.5


Solution Method:
\(f(x)=A\sin(B(x- C))+ D\quad \) where
            A = amplitude
            B = \(\frac{2\pi}{\text{period}}\)
            C = phase(horizontal) shift
            D = vertical shift
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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