Let f(x)=3⋅x4+x3+x2+1x2+x−2. Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity.
used latex on this one
+2446 Asymptotes are always expressed as equations because they are lines, so I would assume that the horizontal asymptote lies at y=0.
Of course, horizontal asymptotes have many rules associated with them. In this case, we are concerned with the following rule: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptotes exists at the x-axis (also known as y=0).
If we take f(x)=3(x4+x3+x2+1)x2+x−2 and multiply it by 1x3, then we would have finished this problem. Therefore, let's find g(x) that does this.
| 3(x4+x3+x2+1)x2+x−2+g(x)=3(x4+x3+x2+1)x3(x2+x−2) | Let's do some subtraction here. |
| g(x)=3(x4+x3+x2+1)x3(x2+x−2)−3(x4+x3+x2+1)(x2+x−2) | Let's convert the rightmost fraction into one with a common denominator so that further simplification is possible. |
| g(x)=3(x4+x3+x2+1)x3(x2+x−2)−3x3(x4+x3+x2+1)x3(x2+x−2) | Let's do some distributing. |
| g(x)=3x4+3x3+3x2+3x3(x2+x−2)−(3x7+3x6+3x5+3x3)x3(x2+x−2) | Now, subtract! |
| g(x)=−3x7−3x6−3x5+3x4+3x2+3x3(x2+x−2) | Not a lot of simplification happens, though, but let's do it anyway! |
You should see that, when adding g(x) to f(x), this results in a horizontal asymptote of y=0.
+2446 Well, I guess I committed a cardinal sin: I did not factor completely. I thought there was no way that either the numerator or denominator could have a common factor. How wrong I was!
| −3x7−3x6−3x5+3x4+3x2+3 | First, factor out a common factor of -3 from every term. |
| −3(x7+x6+x5−x4−x2−1) | Well, I can use the rational root theorem to determine that 1 is a root of the seven-degree polynomial, so x−1 must also be a factor. This results in the following result. After using synthetic division, I got the following! |
| −3(x−1)(x6+2x5+3x4+2x3+2x2+x+1) | Let's worry about the denominator. |
| x3(x2+x−2) | I was able to use factoring here, too, by finding the products of -2 that equal 1: 2 and -1. Let's complete the factoring, then! |
| x3(x−1)(x+2) | |
| −3(x−1)(x6+2x5+3x4+2x3+2x2+x+1)x3(x−1)(x+2) | Look at that! x−1 is a common factor. That's crazy! This leaves us with the following! |
| g(x)=−3(x6+2x5+3x4+2x3+2x2+x+1)x3(x+2),x(x+2)≠0 and x≠1 | I guess this is better! |
