Let f ( x ) = (x^2+8x+12)/(x^2+7x+12).Consider the end behavior and the behavior at each asymptote.

Question

As x → ∞, y→ _____

As x → −∞, y→ _____

As x → −4^-,y→ _____

As x → −4^+,y→ _____

As x → −4^−,y→ _____

As x → −4^+,y→ _____

Answer 1

As both the numerator and denominator are second degree polynomials, the leading coefficient of each is 1, and the numerator and denominator are different, there will be a horizontal asymptote of 1/1 = 1. This applies to x going to both infinity and - infinity.

f(x) = (x+6)(x+2)/((x+3)(x+4))

The zeroes of the numerator are at -6 and -2, while the zeroes of the denominator are at -4 and -3. The zeroes of the numerator are distinct from the zeroes of the denominator. All zeroes are single.

Thus, we can divide the region into (-∞, -6),-6,(-6, -4) , -4, (-4, -3), -3, (-3, -2), -2, (-2, ∞)

As all zeroes are single, the sign of the function alternates. Thus, in the same way that we determined that the horizontal asymptote is 1, the function is positive on (-2, ∞), negative on (-3, -2), positive on (-4, -3), negative on (-6, -4), and positive on (-∞, -6).

Then, while the function has a value of 0 at -6 and -2, it has vertical asymptotes at -4 and -3. The function is positive in the interval (-4, -3) and negative in the adjacent intervals.

Thus, as x -> -4^-, y -> -∞; as x -> -4^+, y -> ∞; as x -> -3^-, y -> ∞; as x -> -3^+, y -> -∞