Math Problem Statement

a) What is the period? Show your work just a subtraction here. ( To find “b” below )

b) What is the Amplitude? VERY Quickly explain how you found it.

c) What is the phase shift? (how much and in what direction. 5 key points are shown)

d) Using above info write an equation to this sine curve in the following format:

a) What is the period? Show your work just a subtraction here. (To find “b” below) b) What is the Amplitude? VERY Quickly explain how you found it. c) What is the phase shift? (how much and in what direction. 5 key points are shown) d) Using the above info write an equation to this sine curve in the following format:

Solution

Let's work through each part of the question based on the provided graph.

a) What is the period?

The period of a sine function can be calculated by looking at the distance between two successive points where the function repeats. In this case, the function completes a full cycle between π6\frac{-\pi}{6} and 5π6\frac{5\pi}{6}.

  • Period = 5π6π6\frac{5\pi}{6} - \frac{-\pi}{6}
  • Period = 5π6+π6=6π6=π\frac{5\pi}{6} + \frac{\pi}{6} = \frac{6\pi}{6} = \pi

Thus, the period is π\pi.

b) What is the amplitude?

The amplitude is the distance from the centerline (which is the x-axis here) to the peak of the sine wave. From the graph, the maximum value of the curve is 2, and the minimum value is -2. The amplitude is the absolute value of this maximum or minimum value.

So, the amplitude is 2.

c) What is the phase shift?

The phase shift is the horizontal shift from the standard sine function. A standard sine function y=sin(x)y = \sin(x) has its first peak at π2\frac{\pi}{2}. In this graph, the first peak occurs at π12\frac{\pi}{12}.

To find the phase shift:

  • Standard peak point = π2\frac{\pi}{2}
  • Peak on the graph = π12\frac{\pi}{12}

Phase shift = π2π12=6π12π12=5π12\frac{\pi}{2} - \frac{\pi}{12} = \frac{6\pi}{12} - \frac{\pi}{12} = \frac{5\pi}{12}

So, the phase shift is 5π12\frac{5\pi}{12} to the right.

d) Write an equation for this sine curve

Using the general form of the sine function: y=Asin(B(xC))+Dy = A \sin(B(x - C)) + D Where:

  • AA is the amplitude
  • BB is related to the period by B=2πPeriodB = \frac{2\pi}{\text{Period}}
  • CC is the phase shift
  • DD is the vertical shift (which is 0 here as the graph oscillates around the x-axis)
  1. Amplitude A=2A = 2.
  2. Period π\pi, so B=2ππ=2B = \frac{2\pi}{\pi} = 2.
  3. Phase shift C=5π12C = \frac{5\pi}{12}.
  4. Vertical shift D=0D = 0.

Thus, the equation is: y=2sin(2(x5π12))y = 2 \sin \left( 2 \left( x - \frac{5\pi}{12} \right) \right)


Would you like to go into more details or have any questions?

Here are 5 related questions to expand your understanding:

  1. How would the equation change if the graph had a vertical shift?
  2. What would the graph look like if the amplitude was 1 instead of 2?
  3. How can we determine the phase shift from key points other than the peak?
  4. What is the relationship between period and frequency in sinusoidal functions?
  5. How would the graph change if the period doubled?

Tip: Always double-check whether the vertical shift (DD) exists by observing if the graph oscillates around a shifted horizontal line.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Functions
Phase Shift
Amplitude
Period

Formulas

Period = 2π / |B|
Amplitude = |A|
Phase Shift = C
y = A sin(B(x - C)) + D

Theorems

Properties of Sine Functions

Suitable Grade Level

Grades 10-12