How do you find the vertical, horizontal or slant asymptotes for #f(x)= -1/(x+1)^2#?

Answer 1

Benjamin Achtenberg

#x=-1# Vertical Asymptote
#y=0# Horizontal Asymptote
No Slant Asymptote

y approaches zero when x approaches infinity therefore #y=0# is a horizontal asymptote.

x approaches -1 when y approaches infinity therefore #x=-1# is a vertical asymptote

#(x+1)^2=-1/y" " "#when y is so large, the right sides of the equation is almost zero and

#x+1=0#

God bless...I hope the explanation is useful..

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Elijah Allen

To find the vertical asymptote(s), set the denominator equal to zero and solve for $x$ . In this case, $(x + 1)^2 = 0$ . Solving for $x$ , we get $x = -1$ . Therefore, the vertical asymptote is $x = -1$ .

To find the horizontal asymptote, examine the behavior of the function as $x$ approaches positive or negative infinity. Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is at $y = 0$ .

There are no slant asymptotes for this function since the degree of the numerator is less than the degree of the denominator.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided 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.