Which of the following is an asymptote of y = sec(x)?
2 answers:
Answer: option d. x = 3π/2
Solution:
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).
Solution:
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).
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Answer:
Step-by-step explanation:
This is a geometric mean problem...a ratio. The geometric mean is the height of the triangle, 6. To solve:
and cross multiply to solve for x:
3x = 36 so
x = 12