h(x) = 5x^2 - 2x - 2 / 2x^2 - 5x + 7 Solution An Asymptote exists if, the functi

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An Asymptote exists if, the function wont be able to return a value for such an input. In other words, the graph of the funciton meets y-axix at infinity or in reality, it like a horizon which feels like kissing the earth, but in reality its not. Hope you have understood. For Horizontal Asymtote: h(x) can be rewritten as: x^2*[5 - 2/x - 2/x^2]/ x^2*[2 - 5/x + 7/x^2] Cancel x^2 on both num and denom and as x-> infinity, 2/x,2/x^2,5/x,7/x^2 goes to zero Thus, at x = 5/2 , we have a HORIZONTAL asymtote. For Vertical Asymtote, When would h(x) be not definable to get an asymtote for such a value of x => denominator = 0 =>2x^2-5x+7 = 0 ( A QEqn) => x = [5 +/- Sqrt (25 - 4*(2)(7)]/4 = 5+/- Sqrt(-31)]/4 =>Here we wont be getting any real value as Sqrt(negative no.) - does not exist. Thus, at any case the denominator doesnt hold to zero. This means, NO Vertical ASYMTOTES !

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