Find the slant asymptote for y = (2x^2 - x - 8) / (2x + 3)

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neela | High School Teacher | (Level 3) Valedictorian

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y = (2x^2-x-8)/(2x+3).

When 2x+3 = 0, x = -3/2 . Or x= -1.5.

At x = -1.5, the denominator becomes zero but not the numerator. So there is a vertical asympotote is there x= -3. Or x= -3/2 is the vertical asymptote.

We cross multiply y = (2x^2-x-8)/(2x+3).

y(2x+3) = 2x^2-x-8.

We rearrange:

(2xy-2x^2) +(x+3y)-8 = 0.

The highest degree terms 2xy-x^2

Put x = 1 and y= m in  2xy-2x^2.

2*1*m - 2*1^2= 2m-2.We equate this to zero and solve for m.

So m = 1.

The second highest degree term (x+3y).

We put x= 1 and  y= m and find p1(x,y) = x+2y and get P1(m) = 1+2m.

Therefore c = -{P1(m)/P2'(m)  for m = 1}= -(1+2m)/(2m-2)' = -(1+3)/2 = -2.

Therefore c = -2.

Therefore  the required oblique asymptote is y = mx+c. Or y = x-2.

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