Question

How do you find the asymptotes for `f(x)=(5/(x-7)) +6`?

Answer 1

vertical asymptote at x = 7
horizontal asymptote at y = 6

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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: `x-7=0rArrx=7" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`f(x)=(5/x)/(x/x-7/x)+6=(5/x)/(1-7/x)+6`

as `xto+-oo,f(x)to0/(1-0)+6`

`rArry=6" is the asymptote"`
graph{((5)/(x-7))+6 [-20, 20, -10, 10]}