Indentify all the vertical asymptotes for 0< or equal to x <2pi, y =tan(2x-(pi/4))?

Answer 1

Where does tan x have vertical asymptotes?
At -pi/2, pi/2, 3*pi/2, 5*pi/2 etc.

So this graph you are dealing with must have asymptotes where
2x - pi/4 = -pi/2, pi/2, 3*pi/2 etc.

Add pi/4 and divide by 2, and you should find that only two of the results are within the domain (0 to 2pi) you've been given, so they are the asymptotes you want.

Answer 2

in view that this seems equivalent to a quadratic equation, permit's replace: permit sin(x) = a we've got right here quadratic equation: 2a^2 +3a + a million = 0 element: (2a + a million)(a+a million) = 0 This yields 2 ideas for a: a = -a million/2 or a = -a million Now we replace sin(x) for a: sin(x) = -a million/2 which suggests x = 5/6 pi sin(x) = -a million which suggests x = 3/2 pi in view that 0 < x < 2pi all of us recognize that those 2 ideas are the unique ideas to 2 sin^2(x)+3sin(x)+a million=0. If this condition weren't recent we would have an infinity of ideas

Answer 3

the rules first

if f(x) + p(x)/q(x) is a ratitioal function in ehich p(x) and q(x) have no common factors and asymptoe is a zero of q(x) the denominator x= vertical asymptote of ''f''

so y = tan(2x-pi/40 is really sin(2x- pi/4)'/ c0s(2 x-pi/4)
there are no cmmon factors so take the demoininator and set it equal to zero
solve for x
cos(2x-pi/4)=0
2x-pi/4
2x=pi/4
x=pi/8
the answer in (radians) is vertical asymptote is x= pi/8
and the cycle repeats evey two pi so only that is only the answer which is pi/8.

Answer 4

Short Way

tan(x) ≠ (pi/2) or (3pi/2)

2x - (pi/4) ≠ (pi/2) or (3pi/2)
2x ≠ (3pi/4) or (7pi/4)
x ≠ (3pi/8) or (7pi/8)