precal-finding horizontal and vertical asymptotes?

Question

how do i find the horizontal and vertical asymptotes?
thanks for the help.

(1) / (x^2 + 4x + 4) would the vertical asymp. be -2?

Answer 1

Horizontal asymptotes: If the degree of the denominator is strictly greater than the degree of the numerator, the horizontal asymptote is at zero. If the degree of the denominator is equal to the degree of the numerator, the horizontal asymptote is given by dividing the coefficient of the first term in the numerator by the coefficient of the first term in the denominator (the coefficients on all other terms do not matter). Finally, if the degree of the denominator is less than the degree of the numerator, there is no horizontal asymptote.

In this case, the horizontal asymptote is at zero.

Vertical asymptotes: these occur wherever the denominator has a real root and the numerator doesn't - this is counting multipliicity (e.g. x/x² has an asymptote at zero because, while the numerator and denominator have roots only at zero, the denominator has two roots at zero while the numerator only has one). In this case, you have two roots in the denominator and none in the numerator - and yes, both of them are at -2, so the sole vertical asymptote would be at -2.

Answer 2

Horizontal asymptose will be find by taking a limit x tends to infinity and dividing Numerator and denominator with the highest power of x. in this question y=0 after simplification