The graph of $y = -\sec(3x)$ is a transformation of the graph of $y = \sec(x)$. We can break this transformation into two main steps:

1. Horizontal compression: The factor of $3$ inside the secant function compresses the graph horizontally by a factor of $\frac{1}{3}$. This affects the period of the secant function, reducing it from $2\pi$ (for $y = \sec(x)$) to $\frac{2\pi}{3}$ (for $y = \sec(3x)$).

2. Vertical reflection: The negative sign in front of the secant function reflects the graph over the x-axis.

Transformation of the Point $(3, 2)$:

To find the transformed point from $(3, 2)$ on the graph of $y = -\sec(3x)$, we apply the same transformations to the x and y coordinates.

• Horizontal compression: The x-coordinate is compressed by a factor of $\frac{1}{3}$. Therefore, $x = 3$ becomes: $x' = \frac{3}{3} = 1$

• Vertical reflection: The y-coordinate is reflected across the x-axis, so $y = 2$ becomes: $y' = -2$

Thus, the transformed point is $(1, -2)$.

Final Transformed Graph

The graph of $y = -\sec(3x)$ will appear as a compressed and flipped version of the original secant graph. The transformed point is $(1, -2)$.

Would you like a graph of this transformation, further details on any steps, or explanations on secant function properties?

Here are 5 related questions:

1. How does the period of $y = \sec(3x)$ compare to $y = \sec(x)$?
2. What are the asymptotes for $y = \sec(x)$ and $y = \sec(3x)$?
3. How does a horizontal compression affect the shape of periodic functions?
4. What other transformations could you apply to $y = \sec(x)$?
5. How does the transformation of points change with horizontal or vertical shifts?

Tip: For transformations, always handle horizontal changes (inside the function) and vertical changes (outside the function) separately to avoid confusion!