SOLUTION: How do I work this problem? Give the equation of the horizontal asymptote of the following rational function: f(x)= x^2+x-6/x-4 Thank you.

Question 57300: How do I work this problem?
Give the equation of the horizontal asymptote of the following rational function:
f(x)= x^2+x-6/x-4
Thank you.
Found 2 solutions by venugopalramana, funmath:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
How do I work this problem?
Give the equation of the horizontal asymptote of the following rational function:
f(x)= x^2+x-6/x-4
Y = F(X)= [X^2+3X-2X-6]/(X+4)=[X(X+3)-2(X+3)]/(X+4)=[(X+3)(X-2)/(X-4)]
THERE IS ONLY 1 VERTICL ASYMPTOTE X=4...SINCE Y TENDS TO INFINITY AS X TENDS TO 4.
THERE IS NO HORIZONTAL ASYMPTOTE.SINCE AS X TENDS TO INFINITY Y ALSO TENDS TO INFINITY
HAD IT BEEN Y=(X-4)/(X^2+X-6)
WE WOULD HAVE GOT A HORIZONTAL ASYMPTOTE OF Y = 0 OR THE X AXIS
SINCE AS X TENDS TO INFINITY Y TENDS TO ZERO.
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
I'm not sure what the denominator is and what the numerator is. If I'm interpretting this wrong, let me know. If you're taking Precal or calculus you have to use limits. Let me know if that's what you're in and I'll show you how to do it using limits.
:
Give the equation of the horizontal asymptote of the following rational function:

:
Here's three rules about horizontal asymtotes and rational functions.
1. If the highest power of the variable (degree) on top is less than the highest power of the variable on the bottom, the horizontal asymptote is y=0.
2. If the degree of the top is the same as the degree on the bottom, the horizontal asymptot is y=leading coeffiecient of the top/leading coefficient of the bottom.
3. If the degree of the top is greater than the degree on the bottom, there is no horizontal asymptote.
:
The degree of the top=2 and the degree of the bottom=1, 2 is greater than 1, therefore there is no horizontal asymptote.
Happy Calculating!!!
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