Question

How do you find vertical, horizontal and oblique asymptotes for `(x² - 3x - 7)/(x+3)`?

Answer 1

`"vertical asymptote at "x=-3`
`"oblique asymptote "y=x-6`

Explanation:

`"let "f(x)=(x^2-3x+7)/(x+3)`

The denominator of f(x) cannot be zero as this would make f(x) 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 then it is a vertical asymptote

`"solve "x+3=0rArrx=-3" is the asymptote"`

Since the degree of the numerator is greater than the degree of the denominator there is an oblique asymptote but no horizontal asymptote.

`"dividing numerator by denominator gives"`

`f(x)=x-6+18/(x+3)`

`"as "xto+-oo,f(x)tox-6+0`

`y=x-6" is the asymptote"`
graph{(x^2-3x-7)/(x+3) [-10, 10, -5, 5]}