How do you find vertical, horizontal and oblique asymptotes for  (x² - 3x - 7)/(x+3)?

Jun 26, 2018

$\text{vertical asymptote at } x = - 3$
$\text{oblique asymptote } y = x - 6$

Explanation:

$\text{let } f \left(x\right) = \frac{{x}^{2} - 3 x + 7}{x + 3}$

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 numerator is non-zero for this value then it is a vertical asymptote

$\text{solve "x+3=0rArrx=-3" is the asymptote}$

Since the degree of the numerator is greater than the degree of the denominator there is an oblique asymptote but no horizontal asymptote.

$\text{dividing numerator by denominator gives}$

$f \left(x\right) = x - 6 + \frac{18}{x + 3}$

$\text{as } x \to \pm \infty , f \left(x\right) \to x - 6 + 0$

$y = x - 6 \text{ is the asymptote}$
graph{(x^2-3x-7)/(x+3) [-10, 10, -5, 5]}