Can someone help me find verticle and horizontal asymptotes?

Question

Hi, need some help with finding asymptotes. Thanks.

f(x)= -4x / x^2 + x - 6

Need horizontal and verticle. thanks for your help. im stuck

Answer 1

I am assuming that your equation is
f(x) = -4x/(x^2 + x - 6)

factorize the denominator
f(x) = -4x/(x-2)(x+3)

Since the denominator cannot be 0, so this equation has vertical asymptotes at x =2 and x = -3

Since the degree in numerator is less the degree in denominator then y=0 is the horizontal asymptote of this eq.

Answer 2

Vertical asymptotes can be found when the denominator equals 0.

x^2 + x - 6 = 0
(x-2)(x+3) = 0
x = 2 and x = -3

I'm not sure how to determine the horizontal asymptote, but graphing the equation it looks like y=0

Answer 3

I will assume that the function is
f(x) = -4x/(x^2 + x - 6)
= -4x/(x+3)(x-2)
A vertical asymptote will be when the denominator becomes zero; look at f(x) with the denom factored: asymptotes exist when x = -3,+2 vertically
Horizontal asymptote: when f(x) is constant ie when f(x) = 0, the x-axis

Answer 4

factor out the demoniator and get:
f(x) = -4x/(x+3)(x-2)
wherever the denom is 0, is a vertical assymp. and that is at x=-3 and x=2

the horizontal assymp is when f(x) =0 which occurs when x=0 (the x axis)