How do you find the asymptotes for # f(x) = (2x-2) /( (x-1)(x^2 + x -1))#?

Answer 1

#"vertical asymptotes at "x~~-1.62" and " x~~0.62#

#"horizontal asymptote at " y=0#

#"the first step is to factorise and simplify f(x)"#
#f(x)=(2cancel((x-1)))/(cancel((x-1))(x^2+x-1))=2/(x^2+x-1)#

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+x-1=0" using the quadratic formula"#
#x=(-1+-sqrt(1+4))/2#
#rArrx=-1/2+-1/2sqrt5#
#rArrx~~-1.62" and " x~~0.62" 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)=(2/x^2)/(x^2/x^2+x/x^2-1/x^2)=(2/x^2)/(1+1/x-1/x^2)#
as #xto+-oo,f(x)to0/(1+0-0)#
#rArry=0" is the asymptote"# graph{2/(x^2+x-1) [-10, 10, -5, 5]}
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Answer 2

To find the asymptotes of the function ( f(x) = \frac{2x - 2}{(x - 1)(x^2 + x - 1)} ), we need to identify vertical, horizontal, and oblique (slant) asymptotes separately.

  1. Vertical Asymptotes: Vertical asymptotes occur where the denominator equals zero, but the numerator doesn't. Set each factor of the denominator equal to zero and solve for ( x ). [ x - 1 = 0 \Rightarrow x = 1 ] [ x^2 + x - 1 = 0 ] This quadratic equation does not have real roots, so it has no effect on vertical asymptotes. Thus, the vertical asymptote is ( x = 1 ).

  2. Horizontal Asymptotes: To find horizontal asymptotes, examine the behavior of the function as ( x ) approaches positive or negative infinity. As ( x ) approaches infinity, the highest power term dominates the function. The highest power term in both the numerator and denominator is ( x ). Divide both the numerator and denominator by ( x ) and find the limit as ( x ) approaches infinity: [ \lim_{x \to \infty} \frac{2x - 2}{(x - 1)(x^2 + x - 1)} ] By simplifying, we get: [ \lim_{x \to \infty} \frac{2 - \frac{2}{x}}{(1 - \frac{1}{x})(x + \frac{1}{2} - \frac{1}{2x})} ] As ( x ) approaches infinity, the terms with ( \frac{1}{x} ) approach zero. Thus, the horizontal asymptote is the ratio of the leading coefficients of the numerator and the denominator, which is ( y = \frac{2}{1} = 2 ).

  3. Oblique (Slant) Asymptotes: Oblique asymptotes occur when the degree of the numerator is one more than the degree of the denominator. Since the degree of the numerator is equal to the degree of the denominator in this case, there are no oblique asymptotes.

To summarize:

  • Vertical asymptote: ( x = 1 )
  • Horizontal asymptote: ( y = 2 )
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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.

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.

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.

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.

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