How do you find vertical, horizontal and oblique asymptotes for # f(x)= (x^3+1 )/( x+1)#?
No asymptotes
so no asymptotes
By signing up, you agree to our Terms of Service and Privacy Policy
To find the vertical, horizontal, and oblique asymptotes for the function ( f(x) = \frac{x^3 + 1}{x + 1} ), follow these steps:

Vertical Asymptotes: Vertical asymptotes occur where the denominator of the rational function is equal to zero and the numerator is nonzero. Set the denominator ( x + 1 ) equal to zero and solve for ( x ): [ x + 1 = 0 ] [ x = 1 ] Therefore, there is a vertical asymptote at ( x = 1 ).

Horizontal Asymptotes: Horizontal asymptotes can be determined by examining the degrees of the numerator and denominator polynomials. Since the degree of the numerator is equal to the degree of the denominator (both are 1), there is a horizontal asymptote. Divide the leading term of the numerator by the leading term of the denominator: [ \lim_{{x \to \infty}} \frac{x^3 + 1}{x + 1} = \lim_{{x \to \infty}} \frac{x^3}{x} = \lim_{{x \to \infty}} x^2 = \infty ] Therefore, there is no horizontal asymptote.

Oblique Asymptotes (Slant Asymptotes): Oblique asymptotes occur when the degree of the numerator is one more than the degree of the denominator. Perform polynomial long division or use synthetic division to divide ( x^3 + 1 ) by ( x + 1 ) to find the oblique asymptote: [ \begin{array}{cc} x^2  x + 1 & x \ \hline x^3 + 0x^2 + 0x + 1 & \ \end{array} ] The quotient is ( x^2  x + 1 ). Therefore, the oblique asymptote is the line ( y = x^2  x + 1 ).
In summary:
 Vertical asymptote: ( x = 1 )
 Horizontal asymptote: None
 Oblique asymptote: ( y = x^2  x + 1 )
By signing up, you agree to our Terms of Service and Privacy Policy
To find the vertical asymptote(s), set the denominator equal to zero and solve for (x). In this case, (x + 1 = 0) gives (x = 1). So, there is a vertical asymptote at (x = 1).
To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (1), there is no horizontal asymptote.
To determine if there are oblique asymptotes, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient will be the equation of the oblique asymptote if the degree of the numerator is exactly one greater than the degree of the denominator. In this case, the degrees are equal, so there are no oblique asymptotes.
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
 If #(1, 0)# lies on the graph of #y = f(x)#, what is the point that lies on the graph of #y = f(x + 3)#?
 How do you find the asymptotes for #4^(x5)5#?
 How do you use composition of functions to show that #f(x)=(2+x)/x# and #f^1(x) = 2/(x1)# are inverses?
 Consider the function #f(x)= 9xx^3#. Is this function odd, even, or neither?
 How do you find the inverse of #y =1/logx#?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7