What are the asymptote(s) and hole(s), if any, of f(x) =(x^2-2x+1)/(x*(x-2))?
2 Answers
See brief explanation
Explanation:
To find the vertical asymptotes, set the denominator -
To find the horizontal asymptote divide the leading term of the numerator -
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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
"solve "x(x-2)=0
x=0" and "x=2" 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)=(x^2/x^2-(2x)/x^2+1/x^2)/(x^2/x^2-(2x)/x^2)=(1-2/x+1/x^2)/(1-2/x)
"as "xto+-oo,f(x)to(1-0+0)/(1-0)
y=1" is the asymptote"
"Holes occur when a common factor is cancelled on the"
"numerator/denominator. This is not the case here hence"
"there are no holes"
graph{(x^2-2x+1)/(x(x-2)) [-10, 10, -5, 5]}