Since secant is the inverse of cosine function, when the cosine function has a 'zero' the secant will have a vertical asymptote.
Graphing a cosine with a right shift of pi, the zeros would be at pi and 2pi.
This will make the vertical asymptotes of the secant function be at pi and 2pi.
I hope you understand.
8,3 is QI
9,-2 is QIV
-3.5,-6 is QIII
Let the number be n. Then 5n + 4 = n. simplifying, 4n = -4, and n = -1
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