How to find the asymptotes of #R(x)=(4x)/(x-3)#?

Question

Cora Alexander · Accepted Answer

vertical asymptote at x = 3
horizontal asymptote at y = 4 A vertical asymptote will occur as the denominator of a rational function tends to zero. To find the equation let denominator equal zero. solve : x - 3 = 0 # rArr x = 3 color(black)(" is the equation")# [ A horizontal asymptote will occur as # lim_(x→±∞) f(x) →0 #] If the numerator and denominator of a rational function are of equal degree the then equation of the asymptote can be found by taking the ratio of leading coefficients. Here they are of equal degree , both of degree 1 . equation is # y = 4/1 = 4 # Here is the graph of the function as an illustration. graph{4x/(x-3) [-40, 40, -20, 20]}

Elijah Allen · Answer

undefined To find the asymptotes of the rational function $R(x) = \frac{4x}{x-3}$, you need to consider both the vertical and horizontal asymptotes.

1. Vertical asymptotes occur where the denominator of the function becomes zero. So, set the denominator equal to zero and solve for $x$:
$$x - 3 = 0$$
$$x = 3$$

Therefore, the vertical asymptote is $x = 3$.

2. Horizontal asymptotes can be found by comparing the degrees of the numerator and the denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients of the two polynomials. In this case, since the degree of the numerator (1) is less than the degree of the denominator (1), the horizontal asymptote is at $y = 0$.

So, the asymptotes of the function $R(x) = \frac{4x}{x-3}$ are: 
- Vertical asymptote: $x = 3$ 
- Horizontal asymptote: $y = 0$