Graph y=sin(-2x)

Problem

$y=\sin (-2x)$

Solution

1. Identify the parent function and its properties. The parent function is $y=\sin x$ which has a period of $2\pi$ an amplitude of $1$ and passes through the origin $(0,0)$

2. Apply the odd function property of the sine function. Since $\sin (-\theta )=-\sin \theta$ the equation $y=\sin (-2x)$ can be rewritten as $y=-\sin 2x$

3. Determine the amplitude. The coefficient of the sine function is $-1$ so the amplitude is $|-1|=1$ This means the graph oscillates between $y=1$ and $y=-1$

4. Calculate the period. The coefficient of $x$ is $B=2$ The period $P$ is calculated using the formula $P=\frac{2\pi }{\left|B\right|}$

$P=\frac{2\pi }{2}=\pi$

5. Identify the phase shift and vertical shift. There are no horizontal or vertical translations, so the phase shift is $0$ and the vertical shift is $0$

6. Find key points for one period $[0,\pi ]$ Divide the period into four equal intervals of length $\frac{\pi }{4}$

• At $x=0$ $y=-\sin 0=0$

• At $x=\frac{\pi }{4}$ $y=-\sin \frac{\pi }{2}=-1$

• At $x=\frac{\pi }{2}$ $y=-\sin \pi =0$

• At $x=\frac{3\pi }{4}$ $y=-\sin \frac{3\pi }{2}=1$

• At $x=\pi$ $y=-\sin 2\pi =0$

7. Sketch the graph by plotting these points and drawing a smooth wave. The graph is a sine wave reflected across the x-axis with a period of $\pi$

Final Answer

$y=-\sin 2x$

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