Question

What is the oblique asymptote of g(x)=x^2-3x-5/x+2

Answer 1

For people doing the Edgenuity Assignment

Explanation:

1. A

2. 450

3. B

4. D

5. C

6. B

7. 1, 3, 4

8. 2, 3, 5

9. C

10. A

11. 1, 4, 7

Answer 2

`\boxed{y = x-5}.`

Explanation:

If the oblique asymptote exists, then its equation is:

`y=mx+b,`

where

`m = \lim\limits_(x\to\pm\infty)(g(x))/(x)`

and

`b = \lim\limits_(x\to\pm\infty)(g(x)-mx).`

Computing the limits with L'Hôpital's rule, we get:

`m = \lim\limits_(x\to\pm\infty) (x^2-3x-5)/(x(x+2))=(x^2-3x-5)/(x^2+2x)\overset{(\infty)/(\infty)}{=}\lim\limits_(x\to\pm\infty) (2x-3)/(2x+2) \overset{(\infty)/(\infty)}{=}\lim\limits_(x\to\pm\infty) (2)/(2) = 1.`

And:

`b = \lim\limits_(x\to\pm\infty)\left((x^2-3x-5)/(x+2)-x\right)= \lim\limits_(x\to\pm\infty)(x^2-3x-5-x^2-2x)/(x+2)=\\\\=\lim\limits_(x\to\pm\infty)(-5x-5)/(x+2) \overset{(\infty)/(\infty)}{=} \lim\limits_(x\to\pm\infty)(-5)/(1)= -5.`

So the oblique asymptote is given by:

`\boxed{y = x-5}.`