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Chapter 4: Problem 17

Sketch the graph of the function. (Include two full periods.) $$y=-2 \tan 3 x$$

Short Answer

Expert verified
The graph of the function \(y = -2 \tan 3x\) shows a series of curves, each extending from negative infinity to positive infinity, with vertical asymptotes at \(x = \pm\frac{\pi}{6} + k \frac{\pi}{3}\) and crossing the x-axis at every \(x = k \frac{\pi}{3}\), decreasing from left to right within each period, with a period of \(\frac{\pi}{3}\).

Step by step solution

01

Understand the Basic Tangent Function

The basic tangent function, \(y = \tan x\), has key characteristics to understand. It has vertical asymptotes (where the function is undefined) at \(x = \pm \frac{\pi}{2} + k\pi\) where \(k\) is any integer, and period \(\pi\). The function crosses the origin, and values increase without bound as \(x\) approaches \(\frac{\pi}{2}\) from the left, and decrease without bound as \(x\) approaches \(-\frac{\pi}{2}\) from the right.
02

Adjust for Amplitude and Period

The given function is \(y = -2 \tan 3x\). The '2' in front of the tangent function affects the amplitude. However, the tangent function technically has no amplitude since it ranges from \(-\infty\) to \(+\infty\), though it does affect the rate of increase or decrease. The '3' inside the tangent function changes its period. Normally, the period of \(\tan x\) is \(\pi\), but a factor of '3' inside compresses it by a factor of 3. This makes the period \(\frac{\pi}{3}\). There will be vertical asymptotes at \(x = \pm\frac{\pi}{6} + k \frac{\pi}{3}\), and the function will cross the origin and every point \(x = k \frac{\pi}{3}\), where \(k\) is an integer.
03

Include the Negative Transformation

The negative sign in front of the function indicates a reflection across the x-axis. Normally, the tangent function increases from left to right within each period. Here, it will decrease from left to right.
04

Sketch the Function

Begin sketching the function by drawing vertical asymptotes at \(x = \pm\frac{\pi}{6} + k \frac{\pi}{3}\) for \(k\) in the range necessary to include two periods of the function. Note that the function crosses the x-axis at every \(x = k \frac{\pi}{3}\). Finally, sketch the curve. It decreases within each period, due to the negative sign, and approaches positive infinity as \(x\) approaches each vertical asymptote from the left, and approaches negative infinity as \(x\) approaches each vertical asymptote from the right.

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