Graph y=2sec(2x)

Problem

y=2*sec(2*x)

Solution

1. Identify the parent function and its properties. The function is a transformation of y=sec(x) which has vertical asymptotes where cos(x)=0

2. Determine the amplitude and vertical stretch. The coefficient 2 outside the secant function indicates a vertical stretch by a factor of 2 The local minima of the curves will be at y=2 and the local maxima will be at y=−2

3. Calculate the period of the function. The period of sec(b*x) is (2*π)/|b| Here, b=2 so the period is (2*π)/2=π

4. Find the vertical asymptotes by setting the argument of the secant function equal to the locations where the cosine function is zero.

cos(2*x)=0

2*x=π/2+n*π

x=π/4+(n*π)/2

The asymptotes occur at x=…,−π/4,π/4,(3*π)/4,…

5. Plot key points within one period.
   At x=0 y=2*sec(0)=2*(1)=2
   At x=π/2 y=2*sec(π)=2*(−1)=−2

6. Sketch the graph by drawing the vertical asymptotes and plotting the U-shaped branches that approach the asymptotes from the key points.

Final Answer

y=2*sec(2*x)

Want more problems? Check here!