Rational Function

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x-int(-3,0) y-int(0,-3) Vertical Asymptote x=-2 Oblique Asymptote y=x-1 What is f(x)?I have this so far f(x)=2(-x-3)/(x+2) which includes the Vertical Asymptote, y-int and x-int.The part I'm having trouble is adding in the Oblique Asymptote without changing the Vertical Asymptote, y-int and x-int.

 

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Ans:

Vertical Asymptote x=-2
So, denominator = (x+2)

we find slant asymptote , dividing numerator by denominator .
And dividend = divisor*quotient + remainder . So,
numerator = (x+2)*(x-1) + k [k= remainder]
= x^2 + 2x -x - 2 + k
= x^2 + x - 2 + k

y = (x^2 + x - 2 + k)/(x+2)
it is passing through (-3, 0) and (0 , -3).

0 = [(-3)^2 -3 - 2 +k] /(-3 + 2)
or, 9 - 5 + k = 0
or, k = -4

and,
-3 = (0 + 0 - 2 + k)/(2)
or, - 6= - 2 + k
or, k = - 4

So, the function is , y = f(x) = (x^2 + x - 2 -4)/(x+2) = (x^2 + x -
6)/(x+2) (Ans.)