61 851
Assignments Done
98%
Successfully Done
In May 2018

Answer to Question #3664 in Algebra for tami

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->∞ (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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

tami
19.07.11, 20:48

Can I get a bit more detail step between the three (y)(x) steps? thanks~

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions