Question #3664

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!

Expert's answer

To have a horizontal asymptote at y = 1 we should find such denominator that

lim_{x->∞} (4x^{2 }+ 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 x^{2}, and shouldn't have one of roots 7/4, -3 or should be the same as 4x^{2 }+ 19x + 21, That is

y(x) (4x^{2 }+ 19x + 21)/4x^{2}

y(x) (4x^{2 }+ 19x + 21)/(4x^{2} +6x -1)

y(x) (4x^{2 }+ 19x + 21)/(4x^{2 }+ 19x + 21)

- would have a horizontal asymptote at y = 1.

lim

In such case f(x) should be a polinomial of degree 2 with 4 before x

y(x) (4x

y(x) (4x

y(x) (4x

- would have a horizontal asymptote at y = 1.

## Comments

tami19.07.11, 20:48Can I get a bit more detail step between the three (y)(x) steps? thanks~

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