How do you find the vertical, horizontal or slant asymptotes for y= (x+1)^2 / ((x-1)(x-3))?

May 21, 2016

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

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : (x-1)(x-3) = 0 → x = 1 , x = 3 are the asymptotes.

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , y \to 0$

When the degree of the numerator and denominator are equal , as is the case here (both of degree 2) then the equation is the ratio of leading coefficients.

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

The leading coefficients(coefficient of ${x}^{2}$ terms) are

$y = \frac{1}{1} = 1 \text{ is the equation of asymptote }$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(x^2+2x+1)/(x^2-4x+3) [-14.24, 14.24, -7.12, 7.12]}