Find the vertical and horizontal asymptote of this
Hint there are two HA and one VA
(√16^x6+9)/9x^3+4
The square root sign is suppose to apply to everything in the parenthesis
Im assuming there the 16 and 9 are squared
Yes
Everything in the parenthesis is suppose to be under the square root
I know to find VA you have to set the denominator equal to 0, but when I do that I get a weird decimal. Would that be my answer?
Also HA is using BOBYO BOTNO EATSDC, acronyms. I got one of the HA to be 4/9 I dont know the other one
To find the horizontal and vertical asymptotes of the given function, let's break down the steps:
Step 1: Vertical Asymptote (VA)
To determine the vertical asymptote, we need to find the values of x for which the function approaches infinity or negative infinity.
In the given function: (√(16^(x^6)+9))/(9x^3+4), we can see that the only denominator is (9x^3+4). Therefore, to find the vertical asymptote, set the denominator equal to zero and solve for x:
9x^3 + 4 = 0
However, this equation does not have any real solutions. As a result, there are no vertical asymptotes for the given function.
Step 2: Horizontal Asymptotes (HA)
To determine the horizontal asymptotes, we analyze the behavior of the function as x approaches positive or negative infinity.
For this step, we need to compare the degrees of the numerator and denominator polynomials. In the given function, the numerator has a degree of 6, and the denominator has a degree of 3.
Since the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes.
In conclusion, there are no vertical asymptotes and no horizontal asymptotes for the given function.