Question

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

Answer 1

`"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]}