The given graph shows a function that looks like a secant or cosecant function, based on the vertical asymptotes and the general shape of the curve. Let's analyze it step by step:

Observations:

• There are vertical asymptotes at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$, which is a common feature of the secant or cosecant functions.
• The shape between the asymptotes is an upward-opening curve, which resembles the secant function, as the basic secant function $y = \sec(x)$ has similar behavior.

Function form:

The general form given is either:

• $y = a \sec(bx) + c$, or
• $y = a \csc(bx) + c$.

From the graph:

• The asymptotes indicate a period of $\pi$. In the secant function $y = \sec(bx)$, the period is $\frac{2\pi}{b}$. Since the period here is $\pi$, we set $b = 2$.
• The secant function's basic graph has an upward curve between asymptotes and looks like the one presented. This confirms we are dealing with a secant function.
• The graph is symmetric around the y-axis, and the vertical shift appears to be zero, so $c = 0$.

Now, to determine the amplitude:

• From the graph, the curve reaches a maximum of 10 and a minimum of -10, suggesting $a = 10$.

Final Equation:

The equation of the function is: $y = 10 \sec(2x)$

Would you like more details, or do you have any questions? Here are some related questions to consider:

1. How does the period of a secant function depend on the value of $b$?
2. What are the key differences between secant and cosecant graphs?
3. How do vertical asymptotes help determine the form of trigonometric functions?
4. What role does the amplitude $a$ play in the graph of a secant function?
5. How can we shift the secant function vertically or horizontally?

Tip: For secant functions, remember that their vertical asymptotes occur where the corresponding cosine function equals zero.