State domain, range, period, vertical asymptotes, zeros, symmetry and y-intercept of: y = -2tan(3x + 180°) + 3.

3 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to use the following trigonometric identity, such that:

`tan(3x + 180^o) = (tan 3x + tan180^o)/(1 - tan180^o*tan 3x)`

Since `tan 180^o = 0` yields:

`tan(3x + 180^o) = (tan 3x + 0)/(1 - 0)`

`tan(3x + 180^o) = tan (3x)`

Replacing `tan(3x) ` for `tan(3x + 180^o)` yields:

`y = -2tan(3x + 180^o) + 3 => y = -2tan (3x) + 3`

The period of tangent function `y = tan(ax)` is evaluated using the following formula, such that:

`p = pi/a`

Identifying a = 3 yields:

`p = pi/3`

Hence, the period of tangent function is `p = pi/3.`

You need to evaluate the domain of the function, such that:

`-npi/2 < 3x < npi/2 => -npi/6 < x < npi/6=> x in (-npi/6, npi/6)`

The domain of the given function is `(-npi/6,npi/6)` and the range of tangent function is the real set of numbers R.

You need to evaluate the vertical asymptotes, hence, you need to evaluate the following limit, such that:

`lim_(x->+-pi/6) (-2tan (3x) + 3) = -oo`

Hence, the vertical asymptotes of the graph of function are `x = +-pi/6.`

You need to find the zeroes of the function, hence, you need to solve the equation `-2tan (3x) + 3 = 0` , such that:

`-2tan (3x) + 3 = 0 => -2tan (3x) = - 3 => tan (3x) = 3/2 => 3x = tan^(-1)(3/2) + npi => x = (tan^(-1)(3/2))/3 + npi/3`

Hence, the zeroes of the function are `x = (tan^(-1)(3/2))/3 + npi/3`

You need to find y intercepts, hence, you need to consider `x = 0` , such that:

`f(0) = -2tan (3*0) + 3 => f(0) = 3`

Hence, the graph intercepts y axis at `(0,3)` and the graph is symetric with respect to `(0,3)` .

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function `y = -2*tan(3x + pi) + 3` .

The domain of the function is the set of real numbers except those where `3x + pi = n*pi + pi/2` or `x = (n*pi - pi/2)/3` . The range of the function is the set of real numbers R. Vertical asymptotes lie at points where `x = (n*pi - pi/2)/3`

The periodicity of the function is `pi/3` . The zeros lie at `3x + pi = n*pi` or `x = (n*pi - pi)/3` . The y-intercept is at (0, 3). The graph is symmetric with respect to the point (0, 3)

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

`y=-2tan(3x+180^o)+3`

domain of y is all real numbers except `3x+180^o !=(n*90^o)` , n is an integer.i.e.

`x!=(n/6-1/3)*180^o=(n-2)*30^o`

Range    `(-oo,oo)`

`y=-2tan(3x+180^o)+3=3-2tan(3x)` ,periodicity of tan is `180^o` ,

periodicity of tan(3x) = `tan(180^o)`

`3x=180^o`

`x=180^o/3=60^o`  this is periodicity of y.

vertical asymptotes  , as `x->+-30^o ,y->oo` so ,vertical asymptotes are

`x=+-30^o=+-pi/6`

Zeros,  `3-2tan(3x)=0`

`` `tan(3x)=1.5`

`tan(3x)=tan(56.31^o)`

`3x=180^o*n+56.31^o`

`x=60^o*n+18.77^o` ,  n is an integer.

y-intercept,  y=3  (put x=0)

 

We’ve answered 318,928 questions. We can answer yours, too.

Ask a question