Function: [tex]y=4sin(2x-\pi)+1[/tex]


Quarter Points of [tex]y=sin(x)[/tex]: [tex](0,0)[/tex] [tex](\frac{\pi}{2} ,1)[/tex] [tex](\pi,0)[/tex] [tex](\frac{3\pi}{2},-1)[/tex] [tex](2\pi,0)[/tex]


Using the coordinates of the five quarter points of y = sin x to determine the corresponding quarter points on the graph of the function above.

Respuesta :

Answer:

The quarter points are

pi /2,1

3 pi/4, 5

pi, 1

5 pi/4, -3

3 pi/2, 1

Step-by-step explanation:

We know that in

[tex]y = \sin(x) [/tex]

the coordinates are

0,0

pi/2,1

pi,0

3 pi/2,-1

2,0

Our function however is new which is

[tex]y = 4 \sin(2x - \pi) + 1[/tex]

Our amplitude is 4 so let multiply the y values by 4.

0,0

pi/2, 4

pi,0

3 pi/2, -4

2 pi,0

Our period is pi now. Remeber that for later.

Our phase shift is pi/2 now so we would move this to the right.

Add pi/2 to the x values.

  • pi/2,0
  • pi,4
  • 3pi/2,0
  • 2 pi , -4
  • 5 pi/2, 0.

Translate the y values up 1.

pi/2,1

pi,5

3 pi/2, 1

2 pi,-3

5 pi/2, 1

The quarter points is the period divide by 4. the period is pi so

[tex] \frac{\pi}{4} [/tex]

Keep pi/2, 1 and add pi/4 to the x values and keep the y value.

  • pi/2,1
  • 3 pi/4, 5
  • pi, 1
  • 5 pi/4, -3
  • 3 pi/2, 1