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)` .

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)