rational functions horizontal asymptote?

Question

is have this equation:
(-2x^3 + 7x^2 + 22x - 48) / (2x^2 - 4x - 16)

in order to find the horiz. asymptote, you need the dominant terms in the numerator and denominator, right?
so wouldnt they be -2x^3 and 2x^2??? is this right?
what are the horizontal asymptotes? are there any??

Answer 1

Nope. No horizontal asymptotes.

A rational function should only have a horizontal asymptote when the degree(largest exponent) of the numerator is equal to or less than the degree(largest exponent) of the denominator.

Answer 2

This function does not have a horizontal asymtote because the degree of the numerator is greater than the degree of the denominator. This means the function has a slant asympotote, also called oblique asymptote.

You can find slant asymptote by performing the long division and discard the remainder. The polynomial you have left is the equation of the slant asymptote.

This function has two vertical asymptotes at x= -2 and x= 4, because the denominator is zero at those values.

Answer 3

vertical asymptote is the place y is undefined enable y = f(x)= -2x+2/x as a result y = -2x^2 + 2 = 2( -x^2 -a million ) = 2 (a million+x) (a million-x) as a result y is undefined at x = -a million and x = a million those are vertical asymptotes horizontal asymptotes is fee of y while x procedures infinity : y = -2x^2 + 2 while x is going to infinity +2 and coffecient of x^2 will become negligible as a result y= -x^2 this is y procedures unfavorable infinity : )