Problem 37 - Precalculus Essentials

Question

Graph two periods of the given cosecant or secant function. $$y=-2 \csc \pi x$$

Step-by-Step Solution

Answer

The graph of \(y=-2 \csc \pi x\) is a transformation of the standard cosecant graph, flipped over x-axis, with vertical asymptotes at integral values of x from 0 to 4 and minima at \(x =0.5, 1.5, 2.5, 3.5\) with the y-value of -2.

Step 1: Identify the period and the range of the function

First, identify the period of the function. The standard period of cosecant function (\(y = csc(x)\)) is \(2\pi\). However, in the function, \(y=-2 \csc(\pi x)\), the constant multiplied with \(x\) is \(\pi\). Therefore, the period will be \(2\pi/\pi=2\). The range of the function is \(-\infty\) to \(-2\) union \(2\) to \(\infty\).

Step 2: Identify the phase shifts and vertical shifts

Since there are no additional constants inside or outside the braces, there are no phase shifts and vertical shifts to adjust for. This means the asymptotes of the function will be at the points where the \(sin(\pi x) = 0\), i.e., the graph can be drawn at integer values of x.

Step 3: Draw the asymptotes

Now, draw dotted vertical lines (asymptotes) at \(x = 0, 1, 2, 3, 4\), which are the integer values within two periods of the function.

Step 4: Plot representative points

Plot points at half distance between the asymptotes, i.e., at \(x =0.5, 1.5, 2.5, 3.5\). Since standard graph of \(csc(x)\) has maxima at these points, the negative sign will make these points minima. So, at \(x = 0.5\), \(y = -2csc(\pi*0.5) = -2\), and similarly for other points.

Step 5: Draw the graph

Finally, connect the points using a smooth curve, remembering to flip the standard cosecant graph due to presence of the '-' sign in the function, and extend the curve towards the asymptotes without touching them. Draw this for two complete periods.