# How do you find vertical, horizontal and oblique asymptotes for (x+1)^2 / ((x-1)(x-3))?

##### 1 Answer
Jan 1, 2017

vertical asymptotes at x = 1 , x = 3
horizontal asymptote at y = 1

#### Explanation:

The denominator of the function cannot be zero as this would make the function 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.

solve : $\left(x - 1\right) \left(x - 3\right) = 0 \Rightarrow x = 1 \text{ or } x = 3$

$\Rightarrow x = 1 \text{ and " x=3" are the asymptotes}$

Horizontal asymptotes occur as

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

$f \left(x\right) = \frac{{x}^{2} + 2 x + 1}{{x}^{2} - 4 x + 3}$

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{2 x}{x} ^ 2 + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4 x}{x} ^ 2 + \frac{3}{x} ^ 2} = \frac{1 + \frac{2}{x} + \frac{1}{x} ^ 2}{1 - \frac{4}{x} + \frac{3}{x} ^ 2}$

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

$\Rightarrow y = 1 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2) Hence there are no oblique asymptotes.
graph{((x+1)^2)/((x-1)(x-3)) [-20, 20, -10, 10]}