Question

find an equation for a negative sine function having an amplitude of 3, a period of \\(\frac{2\pi}{3}\\), and a horizontal shift of \\(\frac{\pi}{3}\\) to the left.\\(y = ?\sin(\quadx + \pi)\\)

Explanation:

Step1: Determine the amplitude

The general form of a sine function is \( y = A\sin(B(x - C)) + D \). The amplitude is \( |A| \). Given the amplitude is 3 and the function is negative, so \( A=- 3\) (but wait, let's check the given form \( y = [?]\sin([?]x+\pi) \). Let's rewrite the general form with horizontal shift. The horizontal shift \( C \) is related to the phase shift. The given horizontal shift is \( \frac{\pi}{3} \) to the left, so the phase shift formula: the phase shift is \( -\frac{C}{B} \) (from \( y = A\sin(Bx + C) \), the phase shift is \( -\frac{C}{B} \)). Wait, the given form is \( y = A\sin(Bx + \pi) \). Let's recall the period formula: period \( T=\frac{2\pi}{|B|} \). Given period \( T = \frac{2\pi}{3} \), so \( \frac{2\pi}{|B|}=\frac{2\pi}{3}\), so \( |B| = 3 \), so \( B = 3 \) (since we need to find \( B \) in the equation). Now for the amplitude: the amplitude is \( |A| \), and the function is negative sine, so \( A=-3 \)? Wait, no, wait the given equation is \( y = [?]\sin([?]x+\pi) \). Let's check the amplitude first. The amplitude is the absolute value of the coefficient of sine. Since it's a negative sine function with amplitude 3, the coefficient \( A \) should be - 3? Wait, no, wait the general form of a negative sine function with amplitude \( a \) is \( y=-a\sin(\dots) \), but here the equation is given as \( y = [?]\sin([?]x+\pi) \). Wait, let's re - express the horizontal shift. The horizontal shift to the left by \( \frac{\pi}{3} \) means that in the function \( y = A\sin(B(x+\frac{\pi}{3})) \) (since left shift is \( +\frac{\pi}{3} \) in the argument). Let's expand this: \( y = A\sin(Bx+\frac{B\pi}{3}) \). But the given form is \( y = A\sin(Bx+\pi) \). So we can set \( \frac{B\pi}{3}=\pi \), solving for \( B \): \( \frac{B\pi}{3}=\pi \), divide both sides by \( \pi \): \( \frac{B}{3}=1 \), so \( B = 3 \). Now for the amplitude: the amplitude is \( |A| = 3 \), and since it's a negative sine function, but let's check the given equation. Wait, the equation is \( y = [?]\sin([?]x+\pi) \). Let's confirm the amplitude: the coefficient of sine is the amplitude (since \( \sin \) has amplitude 1, multiplying by \( A \) gives amplitude \( |A| \)). So the amplitude is 3, and since it's a negative sine function? Wait, no, let's check the phase. Wait, the given form has \( +\pi \) in the argument. Let's recall that \( \sin(x+\pi)=-\sin(x) \), because \( \sin(A + B)=\sin A\cos B+\cos A\sin B \), so \( \sin(x+\pi)=\sin x\cos\pi+\cos x\sin\pi=\sin x(- 1)+0=-\sin x \). So the function \( y = A\sin(Bx+\pi)=-A\sin(Bx) \). We need the amplitude to be 3, so \( | - A| = 3\), so \( |A| = 3 \). Also, the period is \( \frac{2\pi}{|B|}=\frac{2\pi}{3} \), so \( |B| = 3 \), so \( B = 3 \). And since the function is a negative sine function, let's see: \( y=-3\sin(3x) \) is a negative sine function with amplitude 3 and period \( \frac{2\pi}{3} \), but the given form is \( y = A\sin(Bx+\pi) \). Let's substitute \( A = 3 \), then \( y = 3\sin(3x+\pi)=3(-\sin(3x))=-3\sin(3x) \), which has amplitude 3, is a negative sine function (since it's \( - 3\sin(3x) \)), period \( \frac{2\pi}{3} \) (because \( B = 3 \)), and let's check the horizontal shift. The phase shift of \( y = 3\sin(3x+\pi) \) is \( -\frac{\pi}{3} \) (since phase shift is \( -\frac{C}{B} \) for \( y = A\sin(Bx + C) \), here \( C=\pi \), \( B = 3 \), so phase shift is \( -\frac{\pi}{3} \), which means a shift of \( \frac{\pi}{3} \) to the left (because phase shift negative means left shift). So that matches the given horizontal shift of \( \frac{\pi}{3} \) to the left. So the first blank (the coefficient of sine) is the amplitude, which is 3 (but wait, the function is negative? Wait, \( y = 3\sin(3x+\pi)=-3\sin(3x) \), which is a negative sine function. So the coefficient \( A = 3 \), and \( B = 3 \).

Step2: Confirm the values

- For the coefficient of \( \sin \): The amplitude is 3, so the first blank is 3.
- For the coefficient of \( x \): We found \( B = 3 \) from the period formula \( T=\frac{2\pi}{|B|}=\frac{2\pi}{3}\implies|B| = 3\implies B = 3 \).

Answer:

The first blank is \( 3 \) and the second blank is \( 3 \), so the equation is \( y = 3\sin(3x+\pi) \). So the first box (the coefficient of \( \sin \)) is \( 3 \) and the second box (the coefficient of \( x \)) is \( 3 \).