How do you find the asymptotes for #f(x) = (x^2+4x-2)/(x^2-x-7)#?

Question

Abigail Allen · Accepted Answer

#y=1#
#x=(-1-sqrt(29))/2#
#x=(-1+sqrt(29))/2# graph{[-10, 10, -5, 5]} (x^2-4x-2)/(x^2-x-7)

Jace Anderson · Answer

undefined To find the asymptotes for the function f(x) = (x^2+4x-2)/(x^2-x-7), we need to determine the vertical and horizontal asymptotes separately.

Vertical asymptotes occur when the denominator of the function becomes zero. To find them, we set the denominator equal to zero and solve for x. In this case, we solve x^2 - x - 7 = 0. Using the quadratic formula, we find that the solutions are x = (-(-1) ± √((-1)^2 - 4(1)(-7))) / (2(1)). Simplifying this expression gives us x = (1 ± √29) / 2. Therefore, the vertical asymptotes are x = (1 + √29) / 2 and x = (1 - √29) / 2.

Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. In this case, both the numerator and denominator have a degree of 2. To determine the horizontal asymptotes, we compare the leading coefficients of the numerator and denominator. Since they are both 1, the horizontal asymptote is y = 1.

In summary, the vertical asymptotes for f(x) = (x^2+4x-2)/(x^2-x-7) are x = (1 + √29) / 2 and x = (1 - √29) / 2, and the horizontal asymptote is y = 1.