Math Grade 11hi this is a lesson on transforming tangents.  State domain, range, period, zeros and y-intercept of the graph below: The graph could not be posted here so please go to this link...

You need to find the domain and the range of the given function `f(x) = 2 tan(9x + 180) + 9` , hence, you need to remember that the domain of the tangent function is the interval `(-pi/2,pi/2)`  and the range is R.

Hence, since the argument of tangent function needs to be in interval `(-pi/2,pi/2)`  yields:

`-pi/2 < 9x + 180 < pi/2`

You need to substitute `pi`  for 180 such that:

`-pi/2 < 9x +pi < pi/2`

You need to subtract `pi`  both sides such that:

`-pi - pi/2 < 9x < pi/2 - pi => -3pi/2 < 9x < -pi/2`

You need to divide by -9 such that:

`3pi/18 > x > pi/18 => x in (pi/18 , pi/6)`

Hence, the domain of the function is the interval `(pi/18 , pi/6)`  and the range is the interval `(f(pi/18),f(pi/6)).`

You may find the period of tangent function checking the coefficient of x, hence, since the coefficient of x is 9, the period of tangent function is `pi/9` .

You need to find the zeroes of the given function, hence, you need to solve the equation `2 tan(9x + 180) + 9 = 0`  such that:

`2 tan(9x + 180) + 9 = 0 => 2 tan(9x + 180)= -9`

`tan(9x + 180) = -9/2 => 9x + 180 = arctan(-9/2) + n*pi`

`9x = -arctan(9/2) + n*pi - 180 => 9x = -arctan(9/2) + n*pi - pi`

`x =-arctan(9/2) + n*pi/9 - pi/9`

You may find y intercept of the graph considering `x = 0`  such that:

`f(0) = 2 tan(0 + 180) + 9 => f(0) = 2 tan 180 + 9 => f(0) = 2 tan pi + 9 =>f(0) = 2*0 + 9 =>f(0) = 9`

Hence, evaluating the domain of the function yields interval `(pi/18 , pi/6)`  and the range yields the interval `(f(pi/18),f(pi/6)), ` then, evaluating the period of tangent function yields `pi/9,`  the zeroes of the given function yields  `x = -arctan(9/2) + n*pi/9 - pi/9`  and y intercept yields `f(0) = 9` .