How do you identify all asymptotes or holes for #g(x)=(x^2-6x+8)/(x+2)#?

1 Answer
Sep 16, 2017

Answer:

#g(x)# has a vertical asymptote #x=-2# and an oblique (slant) asymptote #y = x-8#

Explanation:

Given:

#g(x) = (x^2-6x+8)/(x+2)#

Divide the numerator by the denominator by grouping:

#g(x) = (x^2-6x+8)/(x+2)#

#color(white)(g(x)) = (x^2+2x-8x-16+24)/(x+2)#

#color(white)(g(x)) = (x(x+2)-8(x+2)+24)/(x+2)#

#color(white)(g(x)) = ((x-8)(x+2)+24)/(x+2)#

#color(white)(g(x)) = x-8+24/(x+2)#

From this alternative expression we can see that #g(x)# has a vertical asymptote at #x=-2# and an oblique (a.k.a. slant) asymptote #y = x-8#.

graph{(y-(x^2-6x+8)/(x+2))(y-x+8) = 0 [-79.6, 80.4, -45.64, 34.36]}