graph $h(x) = 2\\sin(\\pi x - 3\\pi) - 4$ in the interactive widget. note that one moveable point always…

Explanation:

Step1: Simplify the function

First, we can use the trigonometric identity ( \sin(A - B)=\sin A\cos B-\cos A\sin B ), but here we can factor out ( \pi ) from the argument of the sine function: ( h(x) = 2\sin(\pi(x - 3))-4 ). The general form of a sine function is ( y = A\sin(B(x - C))+D ), where:

• ( A ) is the amplitude,
• ( B ) affects the period (( \text{Period}=\frac{2\pi}{|B|} )),
• ( C ) is the phase shift,
• ( D ) is the vertical shift (midline).

For ( h(x)=2\sin(\pi(x - 3))-4 ), we have:

• Amplitude ( |A| = 2 ),
• ( B=\pi ), so the period ( \text{Period}=\frac{2\pi}{\pi}=2 ),
• Phase shift ( C = 3 ) (shift 3 units to the right),
• Vertical shift ( D=- 4 ) (midline is ( y=-4 )).

Step2: Find key points

The standard sine function ( y = \sin(x) ) has key points at ( x = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi ) with corresponding ( y )-values ( 0,1,0, - 1,0 ). For our function ( h(x) ):

1. Midline intersection (neighboring to extremum):

   • The midline is ( y=-4 ). The intersection points with the midline occur when ( \sin(\pi(x - 3)) = 0 ). Solving ( \pi(x - 3)=k\pi ), ( k\in\mathbb{Z} ), we get ( x-3 = k ), so ( x=k + 3 ). For ( k = 0 ), ( x = 3 ), and ( h(3)=2\sin(0)-4=-4 ). Another one, for ( k = 1 ), ( x=4 ), ( h(4)=2\sin(\pi)-4=-4 ).

2. Extremum points:

   • The maximum value of ( \sin(\pi(x - 3)) ) is ( 1 ), so the maximum of ( h(x) ) is ( 2\times1-4=-2 ). This occurs when ( \pi(x - 3)=\frac{\pi}{2}+2k\pi ), ( k\in\mathbb{Z} ). Solving for ( x ): ( x - 3=\frac{1}{2}+2k ), so ( x=\frac{1}{2}+2k + 3=\frac{7}{2}+2k ). For ( k = 0 ), ( x=\frac{7}{2}=3.5 ), ( h(3.5)=2\times1 - 4=-2 ) (maximum, extremum).
   • The minimum value of ( \sin(\pi(x - 3)) ) is ( - 1 ), so the minimum of ( h(x) ) is ( 2\times(-1)-4=-6 ). This occurs when ( \pi(x - 3)=\frac{3\pi}{2}+2k\pi ), ( k\in\mathbb{Z} ). Solving for ( x ): ( x - 3=\frac{3}{2}+2k ), so ( x=\frac{3}{2}+2k + 3=\frac{9}{2}+2k ). For ( k = 0 ), ( x=\frac{9}{2}=4.5 ), ( h(4.5)=2\times(-1)-4=-6 ) (minimum, extremum).

3. Adjust for phase shift and vertical shift:

   • The original graph in the widget seems to have a different vertical shift, but according to our function, the midline is ( y = - 4 ), amplitude ( 2 ), so the extrema are at ( y=-4\pm2 ) (i.e., ( y=-2 ) and ( y = - 6 )) and the period is ( 2 ).

To graph the function:

• Start with the phase shift: shift the standard sine graph (scaled by amplitude ( 2 ), period ( 2 )) 3 units to the right and 4 units down.
• Plot the midline ( y=-4 ).
• Plot the extremum points (max at ( y=-2 ), min at ( y=-6 )) and the intersection points with the midline (where ( y=-4 )) spaced by a quarter - period (since the period is ( 2 ), a quarter - period is ( 0.5 )).

For example, a neighboring intersection with the midline (to an extremum at ( x = 3.5,y=-2 )) would be at ( x=3,y=-4 ) (since from ( x = 3 ) (midline) to ( x = 3.5 ) (extremum) is a quarter - period (0.5 units), and from ( x = 3.5 ) to ( x = 4 ) (midline) is another quarter - period).

Answer:

To graph ( h(x)=2\sin(\pi x - 3\pi)-4 ):

1. Identify key parameters: Amplitude ( 2 ), period ( 2 ), phase shift ( 3 ) (right), vertical shift ( - 4 ) (midline ( y = - 4 )).
2. Plot midline: Draw the horizontal line ( y=-4 ).
3. Plot extremum points: Maximum points (e.g., ( (3.5,-2) )) and minimum points (e.g., ( (4.5,-6) )) using the amplitude and midline.
4. Plot midline intersections: Points like ( (3,-4) ), ( (4,-4) ) (neighboring to extremum points) using the period and phase shift.
5. Connect points smoothly to form the sine - wave graph with the given amplitude, period, phase shift, and vertical shift.