Answer:
Explanation:
The function is:
[tex]\dfrac{-2x^2+3x+6}{x^2+1}[/tex]
The horizontal asymptote is the value of the function when x grows indefinetly, if that value is a constant.
Find the limit when x approaches ± ∞.
[tex]\lim_{x \to \pm \infty} \dfrac{-2x^2+3x+6}{x^2+1}[/tex]
Divide both numerator and denominator by x²
[tex]\lim_{x \to \pm \infty} \dfrac{-2+3/x+6/x^2}{1+1/x^2}[/tex]
As x approaches ±∞, 3/x, 6/x², and 1/x² approach 0.
Thus the result is:
[tex]\lim_{x \to \pm \infty} \dfrac{-2+3/x+6/x^2}{1+1/x^2}=\dfrac{-2}{1}=-2\\\\\\f(x)=-2\\\\y=-2[/tex]
The horizontal asymptote is y = - 2.
