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

Apr 1, 2017

$\text{horizontal asymptote at } y = \frac{1}{2}$

#### 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 and if the numerator is non-zero for these values then they are vertical asymptotes.

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

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

checking the $\textcolor{b l u e}{\text{discriminant}}$

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - \left(4 \times 2 \times 1\right) = - 7$

Since $\Delta < 0$ there are no real solutions 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 ${x}^{2}$

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

as $x \to \pm \infty , f \left(x\right) \to \frac{1}{2 - 0 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

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