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

Transformations of Trig Graphs Exercise 1

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 \(f(x)=3\sin(2x).\)

Answer: 
Graph of one period of a transformed sine function with a max value of 3 at pi over 5 and a min value of -3 at 3 pi over 4 and the middle value of 0 at 0, pi over 2, and pi


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

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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