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# The figure below shows part of the graph of a sine wave. Which of the following equations describes the sine wave? thanks ?

### 3 Answers

- SlowfingerLv 62 months agoFavorite Answer
(D) y = 2(sin(x/2)) + 1

Edit - reasoning explained (if required)

1° period is 4pi so the argument of sine is divided by 2

2° amplitude between min (y= -1) and max (y=+3) is 4 while sine has amplitude 2 (from -1 to +1) --> value of sine is multiplied by 2

3° the function is oscillating around y=1 ----> the sinusoid is shifted up by 1 unit., so there must be "+1" term included

Solution (D) fits in

Check:

x=0 ----> y(0) = 2(sin(0/2)) + 1 = 1

x = pi --> y(pi) = 2(sin(pi/2)) + 1 = 3

x = 2pi -->y(2pi) = 2(sin(pi)) + 1 = 1

x = 3pi -->y(3pi/2) = 2(sin(3pi/2)) + 1 = -1

x = 4pi -->y(4pi) = 2(sin(2pi)) + 1 = 0

ok

- VamanLv 72 months ago
Just check, y=1 when x=0. This gives B, C, and D are valid solutions. A is not the sine wave that represents the graph. But at x=pi, y=3. This possible for D only. D is represented in the graph.

- ?Lv 72 months ago
// y = A sin [B(x+C)] + D

// where A = amplitude

// period = 2π/B ⇒ B = 2π/period

// C = phase shift

// D = vertical shift

BASE function......................y₀ = sin x

VERTICAL STRETCH

by 2 units (A = 2) ................y₁ = 2 sin x

HORIZONTAL STRETCH

by 2 units

(Period = 4π ⇒ B = ¹⁄₂)........y₂ = 2 sin ˣ⁄₂

No phase shift (C=0)...........y₃ = 2 sin ˣ⁄₂

VERTICAL SHIFT by

1 unit (D = 1)........................y₄ = 2 sin ˣ⁄₂ + 1

So, the equation of the graph is

y = 2 sin ˣ⁄₂ + 1 ................ANS (Option D)