How do you find vertical, horizontal and oblique asymptotes for #f(x) = (7x+1)/(2x-9)#?

2 Answers
Aug 6, 2017

Answer:

Vertical asymptote is at # x = 4.5# and horizontal asymptote at
#y =3.5#

Explanation:

# f(x) = (7x+1)/(2x-9)#

For vertical asymptote denominator should be zero ,

#2x-9=0 :. x =9/2 =4.5#. So vertical asymptote is at # x = 4.5#

Here the degree of both denominator and numerator are same

then we have a horizontal asymptote at y = (numerator's leading

coefficient) / (denominator's leading coefficient) i.e at #y= 7/2# or

#y =3.5#. and no oblique asymptote.

Vertical asymptote is at # x = 4.5# and

horizontal asymptote at #y =3.5#

graph{(7x+1)/(2x-9) [-80, 80, -40, 40]}
[Ans]

Aug 6, 2017

Answer:

#"vertical asymptote at "x=9/2#
#"horizontal asymptote at "y=7/2#

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 " 2x-9=0rArrx=9/2" is the asymptote"#

#"horizontal asymptotes occur as"#

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

#"divide the terms on the numerator/denominator by x"#

#f(x)=((7x)/x+1/x)/((2x)/x-9/x)=(7+1/x)/(2-9/x)#

as #xto+-oo,f(x)to(7+0)/(2-0)#

#rArry=7/2" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(7x+1)/(2x-9) [-20, 20, -10, 10]}