Question

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

Answer 1

Answer:

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

Step-by-step explanation:

If the oblique asymptote exists, then its equation is:

$y=mx+b,$

where

$m = \lim\limits_{x\to\pm\infty}\dfrac{g(x)}{x}$

and

$b = \lim\limits_{x\to\pm\infty}(g(x)-mx).$

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

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

And:

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

So the oblique asymptote is given by:

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

Answer 2

Answer:

For people doing the Edgenuity Assignment

Step-by-step 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