# Math Grade 11 Tangents1) State domain, range, period, vertical asymptotes, zeros, symmetry and y-intercept of: y = -2tan(3x + 180°) + 3. 2) Sketch the graph y = 1/(tan(x)), and state its...

Math Grade 11 Tangents

1) State domain, range, period, vertical asymptotes, zeros, symmetry and y-intercept of:

y = -2tan(3x + 180°) + 3.

2) Sketch the graph y = 1/(tan(x)), and state its properties.

if you cannot sketch the graph and show me, then please just state its properties, such as domain, range, period, vertical asymptotes, zeros, symmetry and y-intercept

i would deeply appreciate if you could answer both questions. thank you.

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You need to evaluate the domain of the given function using the condition that the argument of tangent function belongs to the interval `(-pi/2,pi/2)` such that:

`-pi/2 < 3x + pi < pi/2 > -pi/2 - pi < 3x < pi/2 - pi-3pi/2 < 3x < -pi/2`

Multiplying by -1 yields:

`3pi/2 > 3x > pi/2 => pi/2 > x > pi/6`

**Hence, evaluating the domain of the given function yields `x in (pi/6 , pi/2).` **

**You need to remember that the range of tangent function is R.**

**You need to notice that the function has vertical asymptotes at `x = pi/6` and `x = pi/2` such that:**

`lim_(x->pi/6, x>pi/6) (-2tan(3x + 180^o) + 3) = oo`

`lim_(x->pi/2, x>pi/2) (-2tan(3x + 180^o) + 3) = -oo`

You need to remember that the graph of the function intercepts x axis at y = 0, such that:

`-2tan(3x + 180^o) + 3 = 0 => -2tan(3x + 180^o)= -3`

`tan(3x + 180^o) = 3/2 => 3x + 180^o = arctan(3/2) + n*pi`

`3x = arctan(3/2) + n*pi - pi (180^o = pi)`

`x= arctan(3/2)/3 + n*pi/3 - pi/3`

**Hence, evaluating the general solution to the equation of the given function yields `x = arctan(3/2)/3 + n*pi/3 - pi/3` .**

You need to remember that the graph of the function intercepts y axis at x = 0, such that:

`-2tan(3*0 + 180^o) + 3 =y => y = -2tan 18o^o + 3 = y => y = 3`

**Hence, evaluating the y intercept of the graph of the given function yields `(0,3).` **

You need to find the period of the given tangent function, hence, you need to evaluate the ratio `pi/k` , where k represents the coefficient of the argument x, such that:

`p = pi/3 `

**Hence, evaluating the period of the given function yields `pi/3.` **