2 x + 6 = 10solve for x
2 answers:
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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).
Put into slope intercept form, y=mx+b where m=slope
p=-2q+6
slope is -2
subsitute values for q and get values for p
q=0, p=6
(q,p)
(0,6)
(1,4)
p=q+4
slope is 1
subsitutte
q=0,p=4
(q,p)
(0,4)
(1,5)
below are the graphs
p=-2q+6
slope is -2
subsitute values for q and get values for p
q=0, p=6
(q,p)
(0,6)
(1,4)
p=q+4
slope is 1
subsitutte
q=0,p=4
(q,p)
(0,4)
(1,5)
below are the graphs