Answer to Question #3664 in Algebra for tami

Question

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!

Answer 1

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.