How do you find vertical, horizontal and oblique asymptotes for #(2x-1)/(x^2-7x+3)#?

1 Answer
Feb 25, 2017

Answer:

#"vertical asymptotes at "x≈6.54,x≈0.46#
#"horizontal asymptote at "y=0#

Explanation:

#"let "f(x)=(2x-1)/(x^2-7x+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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve "x^2-7x+3=0#

This quadratic does not factorise so use the #color(blue)"quadratic formula"# to find the roots.

#"here "a=1,b=-7" and "c=3#

#• x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(7+-sqrt(49-12))/2=(7+-sqrt37)/2#

#rArrx=6.54 " or "x=0.46" to 2 decimal places"#

#rArrx=6.54" and "x=0.46" are the asymptotes"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=((2x)/x^2-1/x^2)/(x^2/x^2-(7x)/x^2+3/x^2)=(2/x-1/x^2)/(1-7/x+3/x^2)#

as #xto+-oo,f(x)to(0-0)/(1-0+0)#

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator.This is not the case here (numerator-degree 1, denominator-degree 2) Hence there are no oblique asymptotes.
graph{(2x-1)/(x^2-7x+3) [-10, 10, -5, 5]}