Question

for the function i(x)=3- 4/(2x+1) what are my vertical and horizontal asymptotes?

Answer 1

Can anyone help me here please :)

Answer 2

vertical asymptotes occur when denominator equals 0

--> 2x+1 = 0
---> x = -1/2

horizontal asymptotes have 3 rules:
1) numerator has higher exponent than denominator ---> asymptote does not exist

2) numerator exponent equals denominator exponent --> asymptote = ratio of coefficients

3) denominator has higher exponent ---> asymptote = 0

Here the fraction is 4/(2x+1)
- numerator has no variable (exponent of 0)
-denominator has exponent of 1
So we go with rule (3) and say asymptote = 0

**This only applies to fraction!!**
Since function is 3 - 4/(2x+1) , we are adding 3 to the fraction so we must add 3 to the asymptote (its like shifting it up 3)

Horizontal asymptote ---> y=3