How do you find the vertical, horizontal and slant asymptotes of: #y = (2x^2 - 11)/( x^2 + 9)#?

1 Answer
Jul 7, 2017

Answer:

#"horizontal asymptote at " y=2#

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve " x^2+9=0rArrx^2=-9#

#"there are no real solutions hence no vertical asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),ytoc" ( a constant )"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#y=((2x^2)/x^2-11/x^2)/(x^2/x^2+9/x^2)=(2-11/x^2)/(1+9/x^2)#

as #xto+-oo,yto(2-0)/(1+0)#

#rArry=2" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 2 ) hence there are no slant asymptotes.
graph{(2x^2-11)/(x^2+9) [-10, 10, -5, 5]}