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What is oblique asymptote of graph y = (x^2-6x+m)/(2x-4) if m=9?

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eelfy | Student, Undergraduate | (Level 1) eNoter

Posted July 11, 2013 at 3:36 PM via web

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What is oblique asymptote of graph y = (x^2-6x+m)/(2x-4) if m=9?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 11, 2013 at 4:10 PM (Answer #1)

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The line `y = ax + b` is the oblique asymptote for the function `y = (x^2-6x+m)/(2x-4)` if there exists `a = lim_(x->+-oo) y/x` and `b = lim_(x->+-oo)(y - a*x)` .

You need first to evaluate a, at `m = 9` , such that:

`a = lim_(x->+-oo) (x^2 - 6x + 9)/(x(2x - 4))`

You need to force factor `x^2 ` to numerator and `x` to denominator, such that:

`a = lim_(x->+-oo) (x^2(1 - 6/x + 9/x^2))/(x^2(2 - 4/x))`

Reducing duplicate factors yields:

`a = lim_(x->+-oo) (1 - 6/x + 9/x^2)/(2 - 4/x)`

`a = (1 - lim_(x->+-oo) 6/x + lim_(x->+-oo) 9/x^2)/(2 - lim_(x->+-oo) 4/x)`

`a = (1 - 0 + 0)/(2 - 0) => a = 1/2`

Since there exists `a = 1/2` , you need to test if there exists b, such that:

`b = lim_(x->+-oo) ((x^2 - 6x + 9)/(2x - 4) - (1/2)x)`

`b = lim_(x->+-oo) ((x^2 - 6x + 9)/(2(x - 2)) - x/2)`

Bringing the terms to a common denominator yields:

`b = lim_(x->+-oo) ((x^2 - 6x + 9)/(2(x - 2)) - (x(x - 2))/2)`

`b = lim_(x->+-oo) (x^2 - 6x + 9 - x^2 + 2x)/(2(x - 2))`

Reducing duplicate members yields:

`b = lim_(x->+-oo) ( -4x + 9)/(2x - 4)`

Forcing factor x to numerator and denominator yields:

`b = lim_(x->+-oo) (x(-4 + 9/x))/(x(2 - 4/x))`

Reducing duplicate factors yields:

`b = lim_(x->+-oo) (-4 + 9/x)/(2 - 4/x)`

`b = (-4 + lim_(x->+-oo) 9/x)/(2 - lim_(x->+-oo) 4/x)`

`b = -4/2 => b = -2`

Hence, evaluating the oblique asymptote for the given curve, at `m = 9` , yields `y = (1/2)x - 2.`

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