What are the asymptotes of #f(x)=(4tan(x))/(x^2-3-3x)#?

1 Answer
Jun 13, 2017

In resume: The asymptotes of the function are #x = k*pi/2#, #x = k*-pi/2#, #x = 7.58257569496# and #x = -1.58257569496#.

Explanation:

As we can see on the graph below, #4*tan(x)# does have vertical asymptotes. This is known because the value of #tan(x) -> oo# when #x -> k * pi/2# and #tan(x) -> -oo# when #x-> k* -pi/2#.

Important note: #k# is a positive integer. We can use that because it does apply to any multiple of #pi/2# and #-pi/2#.

graph{4*tan(x) [-10, 10, -5, 5]}

Now, we need to check the cases when #f(x)# does not have a real value.

We know that the denominator of the function cannot be 0, because it would create an indeterminacy. So, we also need to check the cases when it does equals 0:

#ax^2 +bx + c = 0#
#x^2 - 3x - 3 = 0#.
Through Bhaskara's formula, we can find the roots of the function:

#Delta = b^2 - 4ac = (-3)^2 - 4(1)(-3) = 9 + 12 = 21#
#x_1 = -b + sqrt(Delta) = 3 + sqrt(21) = 7.58257569496#
#x_2 = -b - sqrt(Delta) = 3 - sqrt(21) = -1.58257569496#

So, now we know that when #x = 7.58257569496# or
#x = -1.58257569496# we have an indeterminacy, as we can see on the graph below:

graph{(4*tan(x))/(x^2-3x-3) [-22.8, 22.8, -11.4, 11.4]}