How do you find the vertical, horizontal and slant asymptotes of: #(-2x+7)/(4x-3)#?

1 Answer
Aug 24, 2016

Answer:

vertical asymptote at #x=3/4#
horizontal asymptote at #y=-1/2#

Explanation:

Let f(x)#=(-2x+7)/(4x-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: #4x-3=0rArrx=3/4" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=(-(2x)/x+7/x)/((4x)/x-3/x)=(-2+7/x)/(4-3/x)#

as #xto+-oo,f(x)to(-2+0)/(4-0)#

#rArry=-1/2" 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{(-2x+7)/(4x-3) [-10, 10, -5, 5]}