How do you find the horizontal asymptote for #f(x)= (x-3)/ (x^2-3x-10)#?

Question

Caleb Alexander · Accepted Answer

No horizontal asymptote in given case.

There is no horizontal asymptote for #f(x)=(x-3)/(x^2-3x-10)#. graph{(x-3)/(x^2-3x-10) [-10, 10, -5, 5]} Horizontal asymptote is there only if in such expressions, degree of numerator is equal to that of denominator. If degree of numerator is one more than that of denominator, we have an oblique or slanting asymptote. The example of latter is #f(x)=(x^2-3x-10)/(x-3)# as #lim_(x->oo)(x^2-3x-10)/(x-3)=lim_(x->oo)(x-3-10/x)/(1-3/x)# = #x-3# and hence oblique or slanting asymptote of #f(x)=(x^2-3x-10)/(x-3)# is #y=x-3#. If #f(x)=(x^2-3x-10)/(2x^2-3)# we have #lim_(x->oo)(x^2-3x-10)/(2x^2-3)# = #lim_(x->ooo=)(1-3/x-10/x^2)/(2-3/x^2)# = #1/2# graph{(x^2-3x-10)/(2x^2-3) [-10, 10, -5, 5]}

Sadie Ackerman · Answer

undefined To find the horizontal asymptote for the function $ f(x) = \frac{x - 3}{x^2 - 3x - 10} $, we examine the behavior of the function as $ x $ approaches positive or negative infinity.

For rational functions like this one, the horizontal asymptote can be determined by comparing the degrees of the numerator and the denominator polynomial.

If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is at $ y = 0 $.

If the degrees of the numerator and the denominator are equal, then the horizontal asymptote is the ratio of the leading coefficients of the numerator and the denominator.

If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.

In this case, the degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is at $ y = 0 $.

Therefore, the horizontal asymptote for the given function $ f(x) = \frac{x - 3}{x^2 - 3x - 10} $ is $ y = 0 $.