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

Jul 12, 2016

vertical asymptote x = 0
slant asymptote y = x + 3

Explanation:

The denominator cannot be zero as this is undefined. Setting the denominator equal to zero and solving will give us the value that x cannot be and if the numerator is non-zero for this value of x then it is a vertical asymptote.

solve x =0 $\Rightarrow x = 0 \text{ is the asymptote}$

When the degree of the numerator > degree of denominator as is the case here (numerator-degree 2, denominator-degree 1) Then we have a slant asymptote but no horizontal asymptote.

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

as $x \to \pm \infty , f \left(x\right) \to x + 3 - 0$

$\Rightarrow y = x + 3 \text{ is the asymptote}$
graph{(x^2+3x-4)/x [-10, 10, -5, 5]}