How do you find the Vertical, Horizontal, and Oblique Asymptote given #(2x^2-8)/(x^2+6x+8)#?

Question

Theodore Amidon · Accepted Answer

vertical asymptote x = -4
horizontal asymptote y = 2

The first step is to factorise and simplify. #rArrf(x)=(2(x-2)cancel((x+2)))/(cancel((x+2))(x+4))=(2x-4)/(x+4)# The denominator cannot be zero as this is undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value of x then it is a vertical asymptote. solve: x + 4 = 0 → x = -4 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by x #((2x)/x-4/x)/(x/x+4/x)=(2-4/x)/(1+4/x)# as #xto+-oo,f(x)to(2-0)/(1+0)# #rArry=2" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) Hence there are no oblique asymptotes. graph{(2x-4)/(x+4) [-20, 20, -10, 10]}

Madelyn Anderson · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. For oblique asymptotes, if the degree of the numerator is one greater than the degree of the denominator, perform polynomial long division to find the equation of the oblique asymptote.