watch the video and then solve the problem given below. click here to watch the video. the function graphed…
watch the video and then solve the problem given below. click here to watch the video. 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. determine the equation of the function. an equation of the function shown is y = □.
Answer
Explanation:
Step1: Determine the type of function
The function has vertical - asymptotes at (x = \pm\frac{\pi}{20}) and is symmetric about the y - axis. The secant function (y = a\sec(bx)+c) is an even function and has vertical asymptotes. So, the function is of the form (y=a\sec(bx)+c).
Step2: Find the value of (c)
The mid - line of the function is (y = 1), so (c = 1).
Step3: Find the value of (b)
The period (P) of the secant function (y = a\sec(bx)+c) is given by (P=\frac{2\pi}{b}). The distance between two consecutive vertical asymptotes is (\frac{\pi}{10}), and for (y = a\sec(bx)+c), the distance between consecutive vertical asymptotes is (\frac{\pi}{b}). Since (\frac{\pi}{b}=\frac{\pi}{10}), then (b = 10).
Step4: Find the value of (a)
When (x = 0), (y=a + c). From the graph, when (x = 0), (y=2). Since (c = 1), then (a+1 = 2), so (a = 1).
Answer:
(y=\sec(10x)+1)