What are the asymptotes of f(x)=(4tan(x))/(x^2-3-3x)f(x)=4tan(x)x233x?

1 Answer
Jun 13, 2017

In resume: The asymptotes of the function are x = k*pi/2x=kπ2, x = k*-pi/2x=kπ2, x = 7.58257569496x=7.58257569496 and x = -1.58257569496x=1.58257569496.

Explanation:

As we can see on the graph below, 4*tan(x)4tan(x) does have vertical asymptotes. This is known because the value of tan(x) -> ootan(x) when x -> k * pi/2xkπ2 and tan(x) -> -ootan(x) when x-> k* -pi/2xkπ2.

Important note: kk is a positive integer. We can use that because it does apply to any multiple of pi/2π2 and -pi/2π2.

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

Now, we need to check the cases when f(x)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 = 0ax2+bx+c=0
x^2 - 3x - 3 = 0x23x3=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]}