How do you identify all asymptotes or holes for #h(x)=(2x^2+9x+2)/(2x+3)#?

1 Answer
Aug 30, 2017

Answer:

#"vertical asymptote at "x=-3/2#
#"oblique asymptote "y=x+3#

Explanation:

The denominator of h(x) cannot be zero as tis would make h(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "2x+3=0rArrx=-3/2" is the asymptote"#

Horizontal asymptotes occur when the degree of the numerator #<=# degree of the denominator. This is not the case here hence there are no horizontal asymptotes.

Oblique (slant ) asymptotes occur when the degree of the numerator #># degree of the denominator. This is the case here hence there is an oblique asymptote.

#"dividing out"#

#x(2x+3)+3(2x+3)-7#

#rArrh(x)=(2x^2+9x+2)/(2x+3)=x+3-7/(2x+3)#

#"as "xto+-oo,h(x)tox+3#

#rArry=x+3" is the asymptote"#

Holes occur when there is a cancellation of a common factor on the numerator/denominator. This is not the case here hence there are no holes.
graph{(2x^2+9x+2)/(2x+3) [-10, 10, -5, 5]}