Question

How do you find the vertical, horizontal or slant asymptotes for `[(9x-4) / (3x+2)] +2`?

Answer 1

vertical asymptote at `x=-2/3`
horizontal asymptote at y = 5

Explanation:

The denominator of the rational function cannot be zero as this would make the rational function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: `3x+2=0rArrx=-2/3" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" (a constant)"`

divide terms on numerator/denominator by x

`f(x)=((9x)/x-4/x)/((3x)/x+2/x)+2=(9-4/x)/(3+2/x)+2`

as `xto+-oo,f(x)to(9-0)/(3+0)+2`

`rArry=5" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1 ) Hence there are no slant asymptotes.
graph{(9x-4)/(3x+2)+2 [-20, 20, -10, 10]}