Question

(Please help!) Find the horizontal asymptote of f(x)=-2x^2+3x+6/x^2+1

Answer 1

Answer:

$\large \boxed{\sf \ \ y=-2 \ \ }$

Step-by-step explanation:

Hello,

To guess the end behaviours when x tends to $\infty$ you only take into account the highest terms of polynomial expressions.

So the expression will be equivalent to

$\dfrac{-2x^2}{x^2}=-2$

In other words we can say

$\displaystyle \lim_{x\rightarrow+\infty} \dfrac{-2x^2+3x+6}{x^2+1}=\lim_{x\rightarrow+\infty} \dfrac{-2x^2}{x^2}=-2\\\\\displaystyle \lim_{x\rightarrow-\infty} \dfrac{-2x^2+3x+6}{x^2+1}=\lim_{x\rightarrow-\infty} \dfrac{-2x^2}{x^2}=-2$

So, the correct answer is y = -2

Hope this helps.

Do not hesitate if you need further explanation.

Thank you