Question

1. using this data, choose the correct graph:

$y = \sin 1.5x$
end point intercept = $(0, 0)$ and $(2\pi, 0)$
period$= \dfrac{2\pi}{b} = \dfrac{2\pi}{1.5} = \dfrac{4\pi}{3}$
$x$-intercept$= \dfrac{1}{2} \dfrac{4\pi}{3} = 2\dfrac{\pi}{3}$ $(\dfrac{2\pi}{3}, \dfrac{4\pi}{3}, 2\pi)$
$a = 1$

Explanation:

Step1: Recall the formula for the period of a sine function

The general form of a sine function is \( y = a\sin(bx) \), and the period is given by \( \text{Period} = \frac{2\pi}{b} \). For the function \( y=\sin(1.5x) \), we have \( b = 1.5=\frac{3}{2} \).

Step2: Calculate the period

Substitute \( b=\frac{3}{2} \) into the period formula: \( \text{Period}=\frac{2\pi}{b}=\frac{2\pi}{\frac{3}{2}} \). When dividing by a fraction, we multiply by its reciprocal, so \( \frac{2\pi}{\frac{3}{2}}=2\pi\times\frac{2}{3}=\frac{4\pi}{3} \).

Step3: Find the x - intercepts

For the sine function \( y = \sin(bx) \), the x - intercepts occur when \( bx = n\pi \), where \( n\in\mathbb{Z} \). For \( y=\sin(1.5x) \), we set \( 1.5x=n\pi \), or \( x=\frac{n\pi}{1.5}=\frac{2n\pi}{3} \). When \( n = 0 \), \( x = 0 \); when \( n = 2 \), \( x=\frac{4\pi}{3} \); when \( n = 3 \), \( x = 2\pi \). Also, the amplitude \( a = 1 \) since the function is \( y=\sin(1.5x) \) (the coefficient of the sine function is 1). The y - intercept is at \( x = 0 \), and \( y=\sin(0)=0 \), so the y - intercept is \( (0,0) \).

Answer:

• Period: \(\boldsymbol{\frac{4\pi}{3}}\)
• x - intercepts: \(\boldsymbol{0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi}\) (and other multiples of \(\frac{2\pi}{3}\))
• y - intercept: \(\boldsymbol{(0,0)}\)
• Amplitude: \(\boldsymbol{1}\)