The problem asks you to graph the trigonometric function:

$y = 3 \sec\left(\frac{3}{2}x\right)$

Steps to sketch the graph:

1. Identify the basic form of the secant function:

   • The secant function is the reciprocal of the cosine function, so its asymptotes occur where the cosine function is zero (since secant is undefined at these points).

2. Determine the period of the secant function:

   • The general form for the secant function is $y = a \sec(bx)$.
   • In this case, $b = \frac{3}{2}$.
   • The period of the secant function is given by $\frac{2\pi}{|b|} = \frac{2\pi}{3/2} = \frac{4\pi}{3}$.

3. Find the asymptotes:

   • The secant function has vertical asymptotes at $x = \frac{\pi}{2} + n\frac{2\pi}{b}$ for any integer $n$, where the cosine function equals zero.
   • For $y = 3\sec\left(\frac{3}{2}x\right)$, the asymptotes occur where $\cos\left(\frac{3}{2}x\right) = 0$, which corresponds to values of $x = \frac{2\pi}{3}n$, for integers $n$.

4. Plot points between asymptotes:

   • The secant function will have a local minimum or maximum midway between asymptotes, depending on the value of $a$. In this case, $a = 3$, so the peaks will be at $y = 3$ and the troughs at $y = -3$.

5. Sketch the graph:

   • Plot three consecutive vertical asymptotes at intervals of $\frac{4\pi}{3}$.
   • Plot points to indicate the maxima and minima of the function between the asymptotes.

Would you like more detailed help on the sketching part, or assistance with any other aspect?

5 related questions:

1. What is the reciprocal relationship between the secant and cosine functions?
2. How do you calculate the period of a trigonometric function like secant?
3. How do vertical asymptotes affect the graph of a secant function?
4. How would the graph change if the amplitude $a$ was negative?
5. Can the secant function ever intersect the x-axis?

Tip:

When sketching secant or cosecant graphs, always start by sketching the cosine or sine graph first, as it helps identify the vertical asymptotes and turning points.