Use the limit theorem and the properties of limits to find the horizontal asymptotes of the graph of the function h(x) = 2x2-6x+1/2+x-x2

Answer:
Option c is answer
Step-by-step explanation:
A function is given as
[tex]h(x) = \frac{2x^2-6x+1}{2+x-x^2}[/tex]
Limit is to be found out for x tends to infinity.
We find that numerator and denominator has the same degree.
HEnce a horizontal asymptote exists
COefficients of leading terms are 2 and -1 respectively
Asymtote would be y =2/-1 = -2
Alternate method:
When x tends to infinity, 1/x tends to 0
[tex]h(x) = \frac{2-\frac{6}{x}+\frac{1}{x^2} }{\frac{2}{x^2} +\frac{1}{x}-1 }[/tex]
by dividing both numerator and denominator by square of x.
Now take limit as 1/x tends to 0
we get
limit is y tends to 2/-1 =-2
Hence horizontal asymptote is y =-2