Answer: option d. x = 3π/2Solution:function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
The answer is: The second one. Why?
Because, when it's a linear graph, the slope of the graph cannot change.
And that's the only graph that fits the criteria
Hope this helped! c:
Answer: 7
Step-by-step explanation:
7
Answer:
2,3,4,5,6,7,8,9,10, that's it easy