How do you find the asymptotes for #y=(x^2+1)/(x^2-9)#?

Question

Madelyn Anderson · Accepted Answer

vertical asymptotes at x = ± 3
horizontal asymptote at y = 1 vertical asymptotes occur when the denominator of a rational function tends to zero. To find equation let denominator = 0 solve # x^2 - 9 = 0 rArr (x+3)(x-3)= 0 rArr x = ± 3# horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0# If the degree of the numerator and denominator are equal, the equation can be found by taking the ratio of leading coefficients. in this case they are both of degree 2 and equation is # y = 1/1 = 1 # here is the graph as an illustration graph{(x^2+1)/(x^2-9) [-10, 10, -5, 5]}

Elias Ackerman · Answer

undefined To find the asymptotes for $y = \frac{x^2 + 1}{x^2 - 9}$, we first identify any vertical asymptotes by determining where the denominator equals zero. In this case, the denominator $x^2 - 9$ equals zero when $x = 3$ and $x = -3$. Therefore, there are vertical asymptotes at $x = 3$ and $x = -3$.

Next, we look for horizontal asymptotes. Horizontal asymptotes occur when the degree of the numerator and denominator are the same. In this case, both the numerator and denominator have a degree of 2. Therefore, we look at the leading coefficients of each term. The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is also 1. When the leading coefficients are equal, the horizontal asymptote is the ratio of the leading coefficients. Thus, the horizontal asymptote is $y = 1$.

Therefore, the asymptotes for the given function are $x = 3$, $x = -3$, and $y = 1$.