How do you find the vertical, horizontal or slant asymptotes for (4x)/(x^2-25)?

Jan 7, 2017

$\text{vertical asymptotes at } x = \pm 5$

$\text{horizontal asymptote at } y = 0$

Explanation:

The denominator of the function cannot be zero as this would make the function 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.

solve : ${x}^{2} - 25 = 0 \Rightarrow {x}^{2} = 25 \Rightarrow x = \pm 5$

$\Rightarrow x = - 5 \text{ and " x=5" are the asymptotes}$

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 ${x}^{2}$

$f \left(x\right) = \frac{\frac{4 x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{25}{x} ^ 2} = \frac{\frac{4}{x}}{1 - \frac{25}{x} ^ 2}$

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

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

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{(4x)/(x^2-25) [-10, 10, -5, 5]}