Which of these rational functions has a ) horizontal asymptote? a slant asymptote? no vertical asymptote?
r(x)= 2x-1/ x^2-x-2 = 2x-1/ (x-2)(x+1)
s(x)= x^3+27/x^2+4 = (x+3)(x^2-3x+9)/(x-2)(x+2)
t(x)=x^3-9x= x(x-3)(x+3)/x+2
u(x)=x^2+x-6/x^2-25= (x+3)(x-2)/(x+5)(x-5)
I do not know how to determine the asymptotes. Please help me. I would sincerely appreciate it. Thanks!
Vertical asymptotes are typically present in functions with a polynomial denominator. The vertical asymptote is located where the value of x is such that the denominator is zero. If the denominator is not a polynomial, the same applies whenever the denominator becomes zero.
To find horizontal asymptotes, you would divide the polynomials by long division.
If the result is of the form p+q/P(x), where p and q are integers and P(x) is a polynomial, the horizontal asymptote is y=p, since the second term vanishes when P(x) becomes infinity.
If the result of the long division is px+q+r/P(x), then the line y=px+q is the oblique asymptote, again, the last term vanishes as x becomes infinity.
You can conclude that the function has no vertical asymptote when the denominator does not vanish for all real values of x.
If you need a check on the answers, feel free to post.
To determine the asymptotes of a rational function, you need to consider its degree and the factors in the numerator and denominator. Here's how you can find the asymptotes for each of the given rational functions:
1. r(x) = (2x - 1) / (x^2 - x - 2)
To find the vertical asymptotes, you need to look for values of x that make the denominator equal to zero. In this case, the denominator can be factored as (x - 2)(x + 1). So, the vertical asymptotes occur at x = 2 and x = -1.
To check for horizontal asymptotes, you compare the degrees of the numerator and denominator. Since the degree of the numerator (1) is less than the degree of the denominator (2), there is a horizontal asymptote at y = 0 (the x-axis).
There are no slant asymptotes for this function.
2. s(x) = (x^3 + 27) / (x^2 + 4)
The denominator, x^2 + 4, has no real roots, so there are no vertical asymptotes.
To check for horizontal asymptotes, compare the degrees of the numerator (3) and denominator (2). Since the degree of the numerator is greater, there is no horizontal asymptote.
To check for slant asymptotes, divide the numerator by the denominator using long division or synthetic division:
(x^3 + 27) / (x^2 + 4) = x + 3 + (9 / (x^2 + 4))
As x approaches positive or negative infinity, the (9 / (x^2 + 4)) term becomes negligible, so the slant asymptote is given by y = x + 3.
3. t(x) = x^3 - 9x
The numerator and denominator have no common factors, so there are no vertical asymptotes.
The degree of the numerator (3) is equal to the degree of the denominator (1), so there is no horizontal asymptote.
Since the degree of the numerator is larger than the degree of the denominator, there is no slant asymptote for this function.
4. u(x) = (x^2 + x - 6) / (x^2 - 25)
The denominator, x^2 - 25, can be factored as (x - 5)(x + 5). So, there are vertical asymptotes at x = 5 and x = -5.
Again, compare the degrees of the numerator (2) and denominator (2) to check for horizontal asymptotes. Since the degrees are equal, divide the leading coefficients: 1/1. Therefore, there is a horizontal asymptote at y = 1.
There are no slant asymptotes for this function.
Remember to always simplify the rational functions and consider any removable discontinuities (holes) that may affect the asymptotes.
To determine the asymptotes of the given rational functions, we need to analyze the behavior of the functions as x approaches positive or negative infinity.
1. r(x) = (2x-1) / (x^2-x-2)
To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Since the degree of the numerator is 1 and the degree of the denominator is 2, r(x) does not have a horizontal asymptote.
To find the vertical asymptotes, we need to find the values of x that make the denominator equal to zero. In this case, the denominator factors as (x-2)(x+1). So, the vertical asymptotes occur at x = 2 and x = -1.
Since r(x) does not have a horizontal asymptote and has two vertical asymptotes, it does not have a slant asymptote.
2. s(x) = (x^3+27) / (x^2+4)
Again, we compare the degrees of the numerator and the denominator. The degree of the numerator is 3, and the degree of the denominator is 2. Therefore, s(x) does not have a horizontal asymptote.
To find the vertical asymptotes, we set the denominator equal to zero: (x-2)(x+2) = 0. So, the vertical asymptotes occur at x = 2 and x = -2.
Since s(x) does not have a horizontal asymptote and has two vertical asymptotes, it does not have a slant asymptote.
3. t(x) = x^3 - 9x
Once again, comparing the degrees of the numerator and the denominator, the degree of both is 3. Therefore, t(x) does not have a horizontal asymptote.
To find the vertical asymptote, we need to find the value of x that makes the denominator equal to zero. In this case, the denominator is x+2. So, the vertical asymptote occurs at x = -2.
Since t(x) does not have a horizontal asymptote and has one vertical asymptote, it does not have a slant asymptote.
4. u(x) = (x^2 + x - 6) / (x^2 - 25)
As before, we compare the degrees of the numerator and the denominator. Both have a degree of 2. Therefore, u(x) does have a horizontal asymptote.
To find the horizontal asymptote, we compare the leading terms of the numerator and the denominator. Both have a leading term of x^2. Therefore, the horizontal asymptote occurs at y = 1.
For the vertical asymptotes, we set the denominator equal to zero: (x+5)(x-5) = 0. So, the vertical asymptotes occur at x = -5 and x = 5.
Since u(x) has a horizontal asymptote, two vertical asymptotes, it does not have a slant asymptote.
In summary:
- r(x) has two vertical asymptotes but no horizontal or slant asymptote.
- s(x) has two vertical asymptotes but no horizontal or slant asymptote.
- t(x) has one vertical asymptote but no horizontal or slant asymptote.
- u(x) has two vertical asymptotes and a horizontal asymptote but no slant asymptote.