Question

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

Answer 1

`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]}