Problem 33 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The graph of \(y=2\sec x\) from \(-2\pi\) to \(2\pi\) consists of a series of U-shaped curves, spaced by vertical asymptotes. The function touches y-axis at \(y=2\) and \(y=-2\) and has vertical asymptotes on either sides of y-axis at \(x=\pi/2\) and \(x=-\pi/2\). The pattern repeats every \(2\pi\).

Step 1: Identify Properties of Function

Identify the key properties of the function \(y=2 \sec x\). This is a secant function with an amplitude of 2, period of \(2\pi\) (same as cosine function) and no phase shift.

Step 2: Plot Critical Points

The critical points of cosine function come at its peak, mid-point and trough i.e. these are the points where \(\cos x = 1, 0, -1\) respectively. For the secant function, these will correspond to points \((x=2n\pi, y=2)\), \((x=(2n+1)\pi/2, y=\text{undefined})\) and \((x=(2n+1)\pi, y=-2)\) where \(n\) is an integer.

Step 3: Draw Vertical Asymptotes

Draw vertical asymptotes at the points where secant function is undefined i.e. \(x = (2n+1)\pi/2\). These lines help to clearly denote where the secant function is undefined.

Step 4: Sketch the Function

Using the critical points and asymptotes, sketch the secant function. The graph will dip downwards at \((2n+1)\pi, -2\) and rise upwards at \((2n+1)\pi, 2\). It will approach the vertical asymptotes and never cross them, since secant function is undefined there.

Step 5: Finish the Graph

Repeat these steps from \(x=-2\pi\) to \(x=2\pi\). This completes the graphing of two periods of the given secant function.