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

Question

Liam Brennan · Accepted Answer

vertical asymptote at x = 3
horizontal asymptote at y = 4

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation equate the denominator to zero. solve : x - 3 = 0 → x = 3 is the equation Horizontal asymptotes occur as #lim_(x →±∞) f(x) → 0# If the degree of the numerator and denominator are equal, as they are here , both of degree 1 , then the equation can be found by taking the ratio of leading coefficients. # y = 4/1 = 4 rArr y = 4 " is the equation "#

Elijah Allen · Answer

undefined To find the vertical, horizontal, or slant asymptotes for the function $ \frac{4x}{x-3} $, we analyze its behavior as $ x $ approaches certain values.

**Vertical Asymptotes:**
Vertical asymptotes occur when the denominator of a rational function becomes zero and the numerator does not. In this case, the denominator $ x - 3 $ becomes zero when $ x = 3 $. Therefore, the function has a vertical asymptote at $ x = 3 $.

**Horizontal Asymptotes:**
To find horizontal asymptotes, we examine the behavior of the function as $ x $ approaches infinity or negative infinity. For a rational function where the degree of the numerator is less than or equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. In this case, the horizontal asymptote is $ y = \frac{4}{1} = 4 $.

**Slant Asymptotes:**
Slant (or oblique) asymptotes occur when the degree of the numerator is one more than the degree of the denominator. To find a slant asymptote, perform polynomial long division or use another method to divide the numerator by the denominator. In this case, the degrees are already such that no slant asymptote exists.

So, for the function $ \frac{4x}{x-3} $:
- It has a vertical asymptote at $ x = 3 $.
- It has a horizontal asymptote at $ y = 4 $.
- It does not have a slant asymptote.