I am trying to figure out this problem: y= 4x^2+19x+21/(Blank)
Write in a denominator that will force this rational function to have a horizontal asymptote at y=1. Some steps shown would be great! Thanks so much!
To have a horizontal asymptote at y = 1 we should find such denominator that lim x->∞ (4x2 + 19x + 21)/f(x) = lim x->∞ (4x - 7)(x + 3)/f(x) = 1 In such case f(x) should be a polinomial of degree 2 with 4 before x2, and shouldn't have one of roots 7/4, -3 or should be the same as 4x2 + 19x + 21, That is y(x) (4x2 + 19x + 21)/4x2 y(x) (4x2 + 19x + 21)/(4x2 +6x -1) y(x) (4x2 + 19x + 21)/(4x2 + 19x + 21) - would have a horizontal asymptote at y = 1.
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I want to sincerely thank you for your help. your work and time that it was completed are more than I could have hoped for and I appreciate your efforts. I look forward to working with you on the next project. which will be in a few months.