How do you find all the asymptotes for function #f(x)= ((3x^16)+28)/ ((15x^9)+33)#?

Question

Sophie Adams · Accepted Answer

#f(x)# has a vertical asymptote #x = root(9)(-11/5)#

It has no horizontal or oblique asymptotes.

#f(x) = (3x^16+28)/(15x^9+33)# #=1/5(15x^16+140)/(15x^9+33)# #=1/5((15x^16+33x^7)-33x^7+140)/(15x^9+33)# #=x^7/5-(33x^7-140)/(15(5x^9+11))# When #x = root(9)(-11/5)# then denominator is zero, but the numerator is non-zero. So there is a vertical asymptote #x = root(9)(-11/5)#. As #x->+-oo#, #(33x^7-140)/(15(5x^9+11)) ->0# since the #9#th power of #x# in the denominator will dominate the #7#th power of #x# in the numerator. So #f(x)# is asymptotic to #x^7/5# as #x->+-oo#. This is not a linear asymptote.

Anna Allen · Answer

undefined To find all the asymptotes of the function $f(x) = \frac{3x^{16} + 28}{15x^9 + 33}$, we need to consider vertical, horizontal, and oblique (slant) asymptotes.

1. Vertical asymptotes occur where the denominator of the rational function is equal to zero, but the numerator is not. For $f(x)$, the denominator is $15x^9 + 33$. Setting this equal to zero gives $15x^9 + 33 = 0$. However, this equation does not have real solutions, so there are no vertical asymptotes.

2. Horizontal asymptotes occur when the degree of the numerator is equal to the degree of the denominator. In this case, the degree of the numerator is 16 and the degree of the denominator is 9. Since the degree of the numerator is greater, there are no horizontal asymptotes.

3. To check for oblique asymptotes, we can perform polynomial division to see if the function can be written as a polynomial plus a proper fraction. Performing the division $3x^{16} + 28 \div (15x^9 + 33)$ yields a quotient that approaches zero as $x$ approaches infinity. Therefore, there is an oblique asymptote at $y = 0$.

In summary, the function $f(x) = \frac{3x^{16} + 28}{15x^9 + 33}$ has an oblique asymptote at $y = 0$, but no vertical or horizontal asymptotes.

Christian Antunez · Answer

undefined To find the asymptotes of the function $f(x) = \frac{{3x^{16} + 28}}{{15x^9 + 33}}$, we look for vertical asymptotes, horizontal asymptotes, and oblique asymptotes.

1. **Vertical Asymptotes**: Vertical asymptotes occur where the denominator of the function approaches zero, but the numerator doesn't. To find vertical asymptotes, we set the denominator equal to zero and solve for $x$.

$$15x^9 + 33 = 0$$
   $$x^9 = -\frac{33}{15}$$
   $$x^9 = -\frac{11}{5}$$

Since the right-hand side is negative, there are no real solutions for $x^9$. Hence, there are no vertical asymptotes.

2. **Horizontal Asymptotes**: Horizontal asymptotes occur when $x$ approaches positive or negative infinity. To find horizontal asymptotes, we compare the degrees of the numerator and denominator.

The degree of the numerator is 16 and the degree of the denominator is 9. Since the degree of the numerator is greater, there are no horizontal asymptotes.

3. **Oblique Asymptotes**: Oblique asymptotes occur when the degree of the numerator is exactly one greater than the degree of the denominator. In this case, the difference in degrees is greater than one, so there are no oblique asymptotes.

Therefore, the function $f(x) = \frac{{3x^{16} + 28}}{{15x^9 + 33}}$ has no vertical, horizontal, or oblique asymptotes.