What are the asymptote(s) and hole(s), if any, of # f(x) =(3x^2)/(5x^2+2x+1) #?

1 Answer
Apr 9, 2017

Answer:

#"horizontal asymptote at " y=3/5#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be.

#"solve " 5x^2+2x+1=0#

This does not factorise hence check #color(blue)"the discriminant"#

#"here " a=5,b=2" and " c=1#

#b^2-4ac=4-20=-16#

Since the discriminant is < 0 there are no real roots hence no vertical asymptotes.

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant )"#

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

#f(x)=((3x^2)/x^2)/((5x^2)/x^2+(2x)/x^2+1/x^2)=3/(5+2/x+1/x^2)#

as #xto+-oo,f(x)to3/(5+0+0)#

#rArry=3/5" is the asymptote"#

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here hence there are no holes.
graph{(3x^2)/(5x^2+2x+1) [-10, 10, -5, 5]}