Question

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

Answer 1

The vertical asymptote is `x=3`
and the slant asymptote is `y=x-6`

Explanation:

As we cannot divide by zero, so `x!=-3`
`:. x=-3` is a vertical asymptote

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. W e have to do a long division.
`color(white)(aaaa)` `x^2-3x-7` `color(white)(aaaa)` `∣` `x+3`
`color(white)(aaaa)` `x^2+3x` `color(white)(aaaaaaaa)` `∣` `x-6`
`color(white)(aaaaa)` `0-6x-7`
`color(white)(aaaaaaa)` `-6x-18`
`color(white)(aaaaaaaaa)` `0+11`

Finally we have, `(x^2-3x-7)/(x-3)=x-6+11/(x+3)`
The slant asymptote is `y=x-6`

graph{(y-((x^2-3x-7)/(x+3)))(y-x+6)=0 [-58.5, 58.5, -29.27, 29.28]}