SOLUTION: {{{ y=(x^2+4x)/(x^3-2x^2-x+2) }}} Find the Horizontal Asymptote. I tried doing this problem: If the numerator and denominator of the equation had the same highest degree, then I

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: {{{ y=(x^2+4x)/(x^3-2x^2-x+2) }}} Find the Horizontal Asymptote. I tried doing this problem: If the numerator and denominator of the equation had the same highest degree, then I      Log On


   

Question 290199: +y=%28x%5E2%2B4x%29%2F%28x%5E3-2x%5E2-x%2B2%29+ Find the Horizontal Asymptote.
I tried doing this problem: If the numerator and denominator of the equation had the same highest degree, then I would just divide the coefficients. What if the equation was bottom "heavy" or the denominator has a higher degree than the numerator? I got confused from here. Please help! Thank you!
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
+y=%28x%5E2%2B4x%29%2F%28x%5E3-2x%5E2-x%2B2%29+
+y=%28x%2A%28x%2B4%29%29%2F%28%28x-2%29+%28x-1%29+%28x%2B1%29%29+

The degree on the denominator (namely, 3) was bigger than the degree on the numerator (namely, 2), and the horizontal asymptote was y = 0 (the x-axis). This property is always true: If the degree on x in the denominator is larger than the degree on x in the numerator, then the denominator, being "stronger", pulls the fraction down to the x-axis when x gets big. That is, if the polynomial in the denominator has a bigger leading exponent than the polynomial in the numerator, then the graph trails along the x-axis at the far right and the far left of the graph.