Question

How do you find the slant asymptote of `y=(x^2+12)/(x-2)`?

Answer 1

The slant asymptote is `y = x+2`

Explanation:

Given: `y = (x^2 + 12)/(x - 2)`

The slant asymptote occurs when the degree in the numerator is one greater than the degree in the denominator.

Since the denominator is a linear factor, you can use either long division or synthetic division to find the slant asymptote.

Using long division:
First set up the divisor and dividend as follows. Make sure the dividend has a term for each degree even if it is zero. Select `x` as the first monomial since `x * x = x^2` Multiply this monomial by each monomial in the divisor:

`" "ul(" "x" ")`
`x - 2|x^2 + 0x + 12`
`" "ul(x^2 -2x)`

Subtract and bring down the next monomial in the dividend:

`" "ul(" "x" ")`
`x - 2|x^2 + 0x + 12`
`" "ul(x^2 -2x)`
`" "2x + 12`

What monomial can we add to the quotient that will eliminate the `2x` term? `" "2`

Multiply this monomial in the quotient by each monomial in the divisor, then subtract:

`" "ul(" "x + 2" ")`
`x - 2|x^2 + 0x + 12`
`" "ul(x^2 -2x)`
`" "2x + 12`
`" "ul(2x -4" ")`
`" "16`

The slant asymptote is `y = x+2`