How do you find vertical, horizontal and oblique asymptotes for #( x + 5) /( 6 - x^3)#?

Question

Anna Allen · Accepted Answer

The vertical asymptote is #x=root(3)6#
The horizontal asymptote is #y=0#
No oblique asymptote

Let #f(x)=(x+5)/(6-x^3)# As you cannot divide by #0#, #x!=root(3)6# The vertical asymptote is #x=root(3)6# As the degree of the numerator is #<# than the degree of the denominator, there is no oblique asymptote #lim_(x->-oo)f(x)=lim_(x->-oo)x/(-x^3)=lim_(x->-oo)-1/x^2=0^-# #lim_(x->+oo)f(x)=lim_(x->+oo)x/(-x^3)=lim_(x->+oo)-1/x^2=0^-# The horizontal asymptote is #y=0# graph{(y-(x+5)/(6-x^3))(y)=0 [-20.27, 20.27, -10.14, 10.14]}

Chloe Alexander · Answer

undefined To find the vertical asymptotes of a rational function, set the denominator equal to zero and solve for $ x $. Any value of $ x $ that makes the denominator zero will be a vertical asymptote.

For the function $ \frac{x + 5}{6 - x^3} $, set the denominator equal to zero:
$$ 6 - x^3 = 0 $$

Solve for $ x $:
$$ x^3 = 6 $$
$$ x = \sqrt[3]{6} $$

So, $ x = \sqrt[3]{6} $ is the vertical asymptote.

To find horizontal and oblique asymptotes, compare the degrees of the numerator and denominator:

1. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $.

2. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.

3. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote, but there might be an oblique asymptote.

In this case, the degree of the numerator is 1, and the degree of the denominator is 3. So, the horizontal asymptote is $ y = 0 $.

There is no oblique asymptote because the degree of the numerator is less than the degree of the denominator.