How do you find the asymptotes for (4x)/(x-3) ?

1 Answer
Apr 24, 2018

"vertical asymptote at "x=3
"horizontal asymptote at "y=4

Explanation:

"let " f(x)=(4x)/(x-3)

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-3=0rArrx=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)=((4x)/x)/(x/x-3/x)=4/(1-3/x)

"as "xto+-oo,f(x)to4/(1-0)

rArry=4" is the asymptote"
graph{(4x)/(x-3) [-20, 20, -10, 10]}