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

1 Answer
Jul 16, 2016

Answer:

vertical asymptote x = -4
horizontal asymptote y = 2

Explanation:

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]}