Question

How do you find the slant asymptote of `f(x) =( x^2 + 3x - 3) / (x+4)`?

Answer 1

The slant asymptote is `y=x-1`

Explanation:

Let's do the long division

`color(white)(aaaa)` `x^2+3x-3` `color(white)(aaaa)` `∣` `color(red)(x+4)`

`color(white)(aaaa)` `x^2+4x` `color(white)(aaaaaaaa)` `∣` `color(blue)(x-1)`

`color(white)(aaaaa)` `0-x-3`

`color(white)(aaaaaaa)` `-x-4`

`color(white)(aaaaaaaaa)` `0+1`

Therefore,

`(x^2+3x-3)/(x+4)=x-1+1/(x+4)`

Let `f(x)=(x^2+3x-3)/(x+4)`

So,

`f(x)=x-1+1/(x+4)`

`lim_(x->-oo)f(x)-(x-1)=lim_(x->-oo)1/x=0^-`

`lim_(x->+oo)f(x)-(x-1)=lim_(x->+oo)1/x=0^+`

The slant asymptote is `y=x-1`

graph{(y-(x^2+3x-3)/(x+4))(y-x+1)=0 [-10, 10, -5, 5]}

Answer 2

Slant asymptote: `y = x-1`. See the asymptotes-inclusive graph.

Explanation:

By division,

`y = f(x)= x-1+1/(x+4)`

Rearranged,

`(y-x+1)(x+4)=1`, revealing that the graph is a hyperbola with

asymptotes

`(y-x+1)(x+4)=0`.

Separately,

y-x+1=0 gives a slant asymptote and

x+4=0 gives a vertical asymptote.

graph{((y-x+1)(x+4)-1)(y-x+1)(x+0.000001y+4)=0 [-23, 23.02, -12.7, 12.6]}