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Question 64619: Find the vertical, horizontal, and oblique asymptotes, if any, as well as any holes, of this rational function: R(x)= (2x^2 - 4x^2 + 3x - 6)/(2x^2 - 3x -2)
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Find the vertical, horizontal, and oblique asymptotes, if any, as well as any holes, of this rational function: R(x)= (2x^2 - 4x^2 + 3x - 6)/(2x^2 - 3x -2)
IS IT 2X^3 OR 2X^2 IN THE NUMERATOR...ASSUMING SO AND USING Y FOR R(X)
N.R. HAS F(2)=0...X-2 IS A FACTOR
N.R = (X-2)(2X^2+3)
D.R.=(2X+1)(X-2)
Y = (X-2)(2X^2+3)/{(X-2)(2X+1)
THIS FUNCTION IS NOT DEFINED AT X=2.
Y TENDS TO INFINITY AS X TENDS TO -1/2....
HENCE X=-1/2 IS A VERTICAL ASYMPTOTE.
AS X TENDS TO INFINITY..Y TENDS TO INFINITY.HENCE THERE IS NO HORIZONTAL ASYMPTOTE.
Y/X = (2X^2+3)(X-2)/(X)(2X+1)(X-2)....TENDS TO 1 = A SAY AS X TENDS TO INFINITY
Y-AX = [(2X^2+3)/(2X+1)]-X = (3-X)/(2X+1)....TENDS TO -1/2 = B SAY AS X TENDS TO INFINITY.
HENCE Y=AX+B = X-0.5 IS THE INCLINED ASYMPTOTE
THERE ARE NO ZEROS FOR THE FUNCTION SINCE 2X^2+3 IS ALWAYS POSITIVE AND THE FUNCTION IS NOT DEFINED AT X=2
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