Give the equation of the horizontal asymptote, if any, of the function. 1) T(x) = 2x^2 - 7x 6/(5x^2 - 9x +6) 2) H(x) = 16x^5 -4/(x-x^3) Please show work, I am trying to understand how to get the asymptote. Solution The horizontal asymptote is what you get when you take x to positive and negative infinity. 1) Divide everything by x^2 giving you: T(x) = (2 - (7/x) + (6/x^2))/(5 - (9/x) + (6/x^2)) When you take x going to positive or negative infinity, all of the terms containing x become zero because x is in the denominator and when the denominator becomes infinitely large, the fraction goes to zero. So you\'re left with 2/5. Thus the horizontal asymptote is y = 5/2. 2) Divide everything by x^5 giving you: H(x) = (16 - (4/x^5))/((1/x^4) - (1/x^2)) When you take x going to negative or positive infinity, we\'re left with only 16 by the same reasoning as before. So the horizontal asymptote is y = 16. .

1. Give the equation of the horizontal asymptote, if any, of the function. 1) T(x) = 2x^2 - 7x
6/(5x^2 - 9x +6) 2) H(x) = 16x^5 -4/(x-x^3) Please show work, I am trying to understand how to
get the asymptote.
Solution
The horizontal asymptote is what you get when you take x to positive and negative infinity. 1)
Divide everything by x^2 giving you: T(x) = (2 - (7/x) + (6/x^2))/(5 - (9/x) + (6/x^2)) When you
take x going to positive or negative infinity, all of the terms containing x become zero because x
is in the denominator and when the denominator becomes infinitely large, the fraction goes to
zero. So you're left with 2/5. Thus the horizontal asymptote is y = 5/2. 2) Divide everything by
x^5 giving you: H(x) = (16 - (4/x^5))/((1/x^4) - (1/x^2)) When you take x going to negative or
positive infinity, we're left with only 16 by the same reasoning as before. So the horizontal
asymptote is y = 16.