How do you find the vertical, horizontal or slant asymptotes for (x^2 + x - 2 )/( x^2 + 3x -4)?

Sep 11, 2016

vertical asymptote at x = - 4
horizontal asymptote at y = 1

Explanation:

The first step is to factorise and simplify f(x).

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

The denominator of f)x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the denominator is non-zero for this value then it is a vertical asymptote.

solve: $x + 4 = 0 \Rightarrow x = - 4 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

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

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(x+2)/(x+4) [-10, 10, -5, 5]}