How do you find vertical, horizontal and oblique asymptotes for (2x-1)/(x^2-7x+3)2x−1x2−7x+3?
1 Answer
Explanation:
"let "f(x)=(2x-1)/(x^2-7x+3)let f(x)=2x−1x2−7x+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=0solve x2−7x+3=0 This quadratic does not factorise so use the
color(blue)"quadratic formula"quadratic formula to find the roots.
"here "a=1,b=-7" and "c=3here a=1,b=−7 and c=3
• x=(-b+-sqrt(b^2-4ac))/(2a)∙x=−b±√b2−4ac2a
x=(7+-sqrt(49-12))/2=(7+-sqrt37)/2x=7±√49−122=7±√372
rArrx=6.54 " or "x=0.46" to 2 decimal places"⇒x=6.54 or x=0.46 to 2 decimal places
rArrx=6.54" and "x=0.46" are the asymptotes"⇒x=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]}
