How do you find the slant asymptote of # ( x^4 + 1 ) / ( x^2 + 2 )#?

Answer 1

This rational function is asymptotic to a parabola, not a line.

It has no slant asymptote.

The degree of the numerator is #4# and the degree of the numerator is #2#.

As a result this rational function is asymptotic to a parabola, not a line.

More explicitly:

#f(x) = (x^4+1)/(x^2+2)#
#=(x^4+2x^2-2x^2-4+5)/(x^2+2)#
#=(x^2(x^2+2)-2(x^2+2)+5)/(x^2+2)#
#=x^2-2 + 5/(x^2+2)#
So as #x->+-oo# we find #(f(x) - (x^2-2)) -> 0#
That is #f(x)# is asymptotic to #x^2-2#
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Answer 2

To find the slant asymptote of the function ( \frac{x^4 + 1}{x^2 + 2} ), we need to perform polynomial long division to divide ( x^4 + 1 ) by ( x^2 + 2 ).

Dividing ( x^4 + 1 ) by ( x^2 + 2 ), we get:

( x^2 - 2x + 4 ) with a remainder of ( -7x + 9 ).

Therefore, the slant asymptote of the function is ( y = x^2 - 2x + 4 ).

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Answer 3

To find the slant asymptote of ( \frac{x^4 + 1}{x^2 + 2} ), perform polynomial long division. The quotient will be the equation of the slant asymptote.

( \frac{x^4 + 1}{x^2 + 2} = x^2 - 2 + \frac{5}{x^2 + 2} )

Therefore, the slant asymptote is ( y = x^2 - 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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