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

1 Answer
Aug 16, 2016

Answer:

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]}