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

Answer 1

vertical asymptotes at x = 1 , x = 3
horizontal asymptote at y = 1

The denominator of the function cannot be zero as this would make the function 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-1)(x-3)=0rArrx=1" or " x=3#

#rArrx=1" and " x=3" are the asymptotes"#

Horizontal asymptotes occur as

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

#f(x)=(x^2+2x+1)/(x^2-4x+3)#

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-(4x)/x^2+3/x^2)=(1+2/x+1/x^2)/(1-4/x+3/x^2)#

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

#rArry=1" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2) Hence there are no oblique asymptotes. graph{((x+1)^2)/((x-1)(x-3)) [-20, 20, -10, 10]}

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Answer 2

Ariana Alvarez

To find the vertical asymptotes, set the denominator equal to zero and solve for $x$ . To find horizontal asymptotes, analyze the degrees of the numerator and denominator polynomials. For oblique asymptotes, perform polynomial long division to divide the numerator by the denominator, and the quotient obtained represents the oblique asymptote.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.