the function graphed to the right is of the form y = a sec bx + c or y = a csc bx + c for some a≠0, b>0…

Explanation:

Step1: Identify the type of function

The graph has vertical asymptotes and U - shaped curves, which is characteristic of a secant function. So the function is of the form $y = a\sec(bx)+c$.

Step2: Find the vertical shift $c$

The mid - line of the graph of $y = a\sec(bx)+c$ is the horizontal line that the graph oscillates around. The mid - line of the given graph is $y = 1$, so $c = 1$.

Step3: Find the period and $b$

The period of $y=\sec(bx)$ is given by $T=\frac{2\pi}{b}$. The distance between two consecutive vertical asymptotes is the period. From the graph, the distance between two consecutive vertical asymptotes is $\pi$. So, $\frac{2\pi}{b}=\pi$. Solving for $b$ gives $b = 2$.

Step4: Find the amplitude $a$

For $y = a\sec(bx)+c$, when $x = 0$, $y=3$. Substituting $x = 0$, $y = 3$, $b = 2$ and $c = 1$ into $y=a\sec(bx)+c$, we get $3=a\sec(0)+1$. Since $\sec(0)=1$, then $3=a\times1 + 1$, which gives $a = 2$.

Answer:

$y = 2\sec(2x)+1$