Question

1. using this data, choose the correct graph:

y = sin 4x
end point intercept = (0, 0) and (2π, 0)
period = \\(\frac{2π}{b}\\) = \\(\frac{2π}{4}\\) = \\(\frac{π}{2}\\)
x-intercept = \\(\frac{1}{2} \cdot \frac{π}{2}\\) = \\(\frac{π}{4}\\)
(\\(\frac{π}{4}\\), \\(\frac{π}{2}\\), \\(\frac{3π}{4}\\), π, ...)
a = 1

Explanation:

Step1: Recall the general form of sine function

The general form of a sine function is \( y = a\sin(bx) \), where \( a \) is the amplitude, the period is \( \frac{2\pi}{b} \), and the \( x \)-intercepts occur where \( \sin(bx)=0 \), i.e., \( bx = n\pi \) for integer \( n \), so \( x=\frac{n\pi}{b} \).

Step2: Analyze the given function \( y = \sin(4x) \)

For the function \( y=\sin(4x) \), we have \( a = 1 \) (amplitude) and \( b = 4 \).

Step2.1: Calculate the period

Using the period formula \( \text{Period}=\frac{2\pi}{b} \), substitute \( b = 4 \):
\( \text{Period}=\frac{2\pi}{4}=\frac{\pi}{2} \)

Step2.2: Find the \( x \)-intercepts

Set \( y = 0 \), so \( \sin(4x)=0 \). This implies \( 4x=n\pi \), where \( n \) is an integer. Solving for \( x \), we get \( x = \frac{n\pi}{4} \). For \( n = 0 \), \( x = 0 \); for \( n = 1 \), \( x=\frac{\pi}{4} \); for \( n = 2 \), \( x=\frac{2\pi}{4}=\frac{\pi}{2} \); for \( n = 3 \), \( x=\frac{3\pi}{4} \); for \( n = 4 \), \( x=\pi \), and so on. So the \( x \)-intercepts are \( 0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\dots \)

Step2.3: End - point intercepts

The function \( y = \sin(4x) \) has a period of \( \frac{\pi}{2} \). If we consider a full period or multiple periods, the end - point intercepts (for example, over an interval) will follow the pattern of the \( x \)-intercepts. At \( x = 0 \), \( y=\sin(0)=0 \), and if we consider the end of a period, say from \( x = 0 \) to \( x=\frac{\pi}{2} \) (one period), the end - point at \( x=\frac{\pi}{2} \) also gives \( y=\sin(4\times\frac{\pi}{2})=\sin(2\pi)=0 \). For a larger interval, like from \( x = 0 \) to \( x = 2\pi \), the number of periods in \( 2\pi \) is \( \frac{2\pi}{\frac{\pi}{2}}=4 \) periods. At \( x = 2\pi \), \( y=\sin(4\times2\pi)=\sin(8\pi)=0 \), so the end - point intercepts (for the interval from \( 0 \) to \( 2\pi \)) are \( (0,0) \) and \( (2\pi,0) \) as the function value at these points is \( 0 \).

To determine the correct graph, we look for a sine - wave with amplitude \( 1 \), period \( \frac{\pi}{2} \), \( x \)-intercepts at \( x=\frac{n\pi}{4} \) ( \( n \) is an integer), and passing through \( (0,0) \) and \( (2\pi,0) \). The graph should have 4 full periods in the interval \( [0,2\pi] \) (since the period is \( \frac{\pi}{2} \) and \( \frac{2\pi}{\frac{\pi}{2}} = 4 \)), with peaks at \( x=\frac{\pi}{8},\frac{5\pi}{8},\frac{9\pi}{8},\frac{13\pi}{8} \) (where \( 4x=\frac{\pi}{2}+2n\pi \), so \( x=\frac{\pi}{8}+\frac{n\pi}{2} \)) and troughs at \( x=\frac{3\pi}{8},\frac{7\pi}{8},\frac{11\pi}{8},\frac{15\pi}{8} \) (where \( 4x=\frac{3\pi}{2}+2n\pi \), so \( x=\frac{3\pi}{8}+\frac{n\pi}{2} \)).

Answer:

The correct graph of \( y = \sin(4x) \) should have an amplitude of \( 1 \), a period of \( \frac{\pi}{2} \), \( x \)-intercepts at \( x=\frac{n\pi}{4} \) ( \( n\in\mathbb{Z} \) ), and pass through \( (0,0) \) and \( (2\pi,0) \) with 4 full cycles in the interval \( [0,2\pi] \).