How do you find the vertical, horizontal or slant asymptotes for #f(x)=( x-4)/ (x^2-1)#?

Question

Elias Ackerman · Accepted Answer

undefined To find the vertical asymptotes of the function f(x) = (x - 4)/(x^2 - 1), you need to identify the values of x for which the denominator is equal to zero. In this case, the denominator x^2 - 1 = 0 when x = ±1. Therefore, the vertical asymptotes are x = 1 and x = -1.

To find horizontal or slant asymptotes, you need to examine the behavior of the function as x approaches positive or negative infinity. Since the degree of the numerator is less than or equal to the degree of the denominator, the horizontal asymptote can be found by dividing the leading coefficients of the numerator and denominator. In this case, the leading coefficients are both 1, so the horizontal asymptote is y = 0 (the x-axis).

Since the degree of the numerator is less than the degree of the denominator by exactly 1 (degree 1 in the numerator and degree 2 in the denominator), there is a slant asymptote. To find it, perform polynomial long division or use other methods to divide the numerator by the denominator. The quotient will represent the equation of the slant asymptote.

Ariana Alvarez · Answer

vertical asymptotes x = ± 1
horizontal asymptote y = 0

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero. solve : #x^2-1=0rArrx^2=1rArrx=±1# #rArrx=-1,x=1" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by #x^2# #(x/x^2-4/x^2)/(x^2/x^2-1/x^2)=(1/x-4/x^2)/(1-1/x^2)# as #xto+-oo,f(x)to(0-0)/(1-0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes. graph{(x-4)/(x^2-1) [-10, 10, -5, 5]}

Caleb Alexander · Answer

undefined To find the vertical asymptotes of a rational function, you need to look for values of $ x $ that make the denominator equal to zero, but the numerator doesn't. For the given function $ f(x) = \frac{x-4}{x^2-1} $, the denominator equals zero when $ x = 1 $ or $ x = -1 $, but the numerator doesn't. Thus, the vertical asymptotes are $ x = 1 $ and $ x = -1 $.

Horizontal asymptotes of a rational function are determined by comparing the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. In this case, both the numerator and the denominator have a degree of 1, so the horizontal asymptote is $ y = \frac{1}{1} = 1 $.

There are no slant asymptotes for this function because the degree of the numerator is less than the degree of the denominator by 1.