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Chapter 4: Problem 10

Graph two periods of the given tangent function. $$y=-3 \tan \frac{1}{2} x$$

Short Answer

Expert verified
The function y=-3 tan(1/2 x) has a period of \(2\pi\), no phase or vertical shift, has a vertical stretch of '3', a reflection over x-axis and x-intercepts at multiples of \(\pi\) and vertical asymptotes at odd multiples of \(\pi/2\). Therefore, the graph first moves from the highest point and goes down the x-axis at \(\pi\), reaches lowest point at \(2\pi\), rises upwards and repeats this pattern.

Step by step solution

01

Identify the Amplitude

For tangent and cotangent functions, it's important to understand that the coefficient, in this case -3, doesn't affect the amplitude, but causes a vertical stretch. It specifically means that the graph will extend \(3\) units above the x-axis before returning to the x-axis.
02

Determine the Reflection

The negative sign in front of the 3 in the given equation indicates a reflection over the x-axis. This means the graph, instead of starting from the origin, will start from the highest point.
03

Identify the Period

The period for any tangent function is \(\pi/B\), where 'B' represents the coefficient of 'x'. In this equation, \(B= \frac{1}{2}\). Hence, the period is \(\pi/\frac{1}{2}\) which equals \(2\pi\). So, the given function completes one full cycle over an interval of \(2\pi\).
04

Identify the Phase Shift and Vertical Shift

Since there's no value added or subtracted inside the function or outside, there is no phase shift or vertical shift. Hence, the graph will start at the origin.
05

Draw the Graph

Instead of starting at \(0\), the function starts at the highest point due to the reflection and then comes down to cross the x-axis at \(\pi\), goes to the lowest point at \(2\pi\) and then starts rising back. The x-intercepts are at \(n\pi\) where n is any integer, and there are vertical asymptotes at \(n\pi + \pi/2\) where 'n' is any integer.

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