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

Apr 9, 2017

$\text{horizontal asymptote at } y = \frac{3}{5}$

#### 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.

$\text{solve } 5 {x}^{2} + 2 x + 1 = 0$

This does not factorise hence check $\textcolor{b l u e}{\text{the discriminant}}$

$\text{here " a=5,b=2" and } c = 1$

${b}^{2} - 4 a c = 4 - 20 = - 16$

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

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant )}$

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

$f \left(x\right) = \frac{\frac{3 {x}^{2}}{x} ^ 2}{\frac{5 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{3}{5 + \frac{2}{x} + \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{3}{5 + 0 + 0}$

$\Rightarrow y = \frac{3}{5} \text{ is the asymptote}$

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]}