State domain, range, period, vertical asymptotes, zeros, symmetry and y-intercept of: y = -2tan(3x + 180°) + 3.
3 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
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 | College Teacher | (Level 2) Distinguished Educator
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 | College Teacher | (Level 3) Valedictorian
`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)