How do you find the vertical, horizontal or slant asymptotes for (x-7)/(3x^2+17x-6 )?

Mar 5, 2017

$\text{vertical asymptotes at "x=-6" and } x = \frac{1}{3}$
$\text{horizontal asymptote at } y = 0$

Explanation:

$\text{let } f \left(x\right) = \frac{x - 7}{{x}^{2} + 17 x - 6}$

The denominator of f(x) cannot be zero as this would make f(x) 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.

$\text{solve: } 3 {x}^{2} + 17 x - 6 = 0 \Rightarrow \left(3 x - 1\right) \left(x + 6\right) = 0$

$\Rightarrow x = - 6 \text{ and "x=1/3" are the asymptotes}$

Horizontal asymptotes occur as

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

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

$f \left(x\right) = \frac{\frac{x}{x} ^ 2 - \frac{7}{x} ^ 2}{\frac{3 {x}^{2}}{x} ^ 2 + \frac{17 x}{x} ^ 2 - \frac{6}{x} ^ 2} = \frac{\frac{1}{x} - \frac{7}{x} ^ 2}{3 + \frac{17}{x} - \frac{6}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x-7)/(3x^2+17x-6) [-10, 10, -5, 5]}