How do you find the slant asymptote of #y=(x^3 - 4x^2 + 2x -5)/ (x^2 + 2)#?

Answer 1

#color(blue)("Slant Asymptote")#
#color(blue)(y=x-4)#

To find the slant asymptote, we divide #x^3-4x^2+2x-5# by #x^2+2#

The resulting quotient not including the remainder part represents the slant asymptote

Let us divide

#" " " " " " " " " " " "underline(x-4" " " " " " " " " " " " " ")#
#x^2+0*x+2|~x^3-4x^2+2x-5#
#" " " " " " " " " " " "underline(x^3+0x^2+2x" " " " " " " ")#
#" " " " " " " " " " " " " "-4x^2+0-5#
#" " " " " " " " " " " " " "underline(-4x^2+0x-8" " " " " ")#
#" " " " " " " " " " " " " " " " " " " " "" " " " +3#

Observe the quotient #x-4# so that our slant asymptote is

#y=x-4#

Kindly see the graph of #y=(x^3-4x^2+2x-5)/(x^2+2)" "#(colored red) and the slant asymptote #y=x-4" "#(colored blue).

God bless....I hope the explanation is useful.

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Answer 2

Caleb Alexander

To find the slant asymptote of ( y = \frac{x^3 - 4x^2 + 2x - 5}{x^2 + 2} ), perform polynomial long division or synthetic division to divide ( x^3 - 4x^2 + 2x - 5 ) by ( x^2 + 2 ). The quotient obtained represents the equation of the slant 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.