How do you sketch the graph of $y = sec (3 x + \frac{π}{2})$ $y = sec(3x + pi/2)$ ?

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Answer 1 · 2018-07-22T11:01:27

See explanation and graph.

$y = sec (3 x + \frac{π}{2}) = - csc 3 x \notin (- 1, 1)$ $y =sec ( 3 x + pi/2 ) = - csc 3x notin ( - 1, 1 )$

and $3 x \neq$ $3x ne$ an integer multiple of $π$ $pi$

$\Rightarrow x \neq$ $rArr x ne$ integer multiple of $\frac{π}{3}$ $pi/3$

The asymptotes are $x = k(pi/3), k = 0, +-1, +-2, +-3,...$

The period = period of sin 3x = (2pi)/3

Rewrite $y sin 3x + 1 = 0$ , by cross multiplication..

For this form, the Socratic graph is immediate.
graph{(ysin(3x) + 1)(y^2-1)(x^2-1.0966)= 0[-10 10 -5 5]}

Observe

$y notin ( - 1, 1 )$ and the asymptotes $x = +- pi/3$ , near O and also the period $(2pi)/3$ .

How do you sketch the graph of y = sec(3x + pi/2)y=sec(3x+π2)y = sec(3x + pi/2)?