How do you graph $y=-2sec(4x+(3pi)/4)+2$ ?

Question

1973 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-18T03:56:13

See graph and explanation.

The period = $(2pi)/4 = pi/2$

$( 2 - y )/2 = sec ( 4x + ( 3pi )/4 ) notin ( -1, 1 )$

$rArr y notin ( 0, 4 )$

The asymptotrd are given by

$4 x + (3pi)/4 = ( 2 k+ 1 ) pi/2$

$rArr x = ( 4 k - 1 )pi/16, k = 0, +-1, +-2, +_3, ...$

graph{(cos(4x+(3pi)/4)(1-y/2)-1)(y)(y-4)(x+(pi/16))(x-(3pi)/16) = 0[-3 3 -6 10]}

The not-to-uniform scale graph includes $y notin ( 0, 4 ) space$

and the two asymptotes near O,

$x = - pi/6 and x = (3pi)/16$

How do you graph y=-2sec(4x+(3pi)/4)+2y=-2sec(4x+(3pi)/4)+2?