Why does the end behavior of a polynomial behave as it does?

[howiejackson14]

Why does the end behavior of a polynomial behave as it does? For example, why does f(x) approach infinity as x approaches infinity and x approaches negative infinity when a polynomial has an even degree with a positive leading coefficient?

Could you explain this to me algebraically and not with calculus reasoning!! Need Help!!!!!

 

[Ishuda]

For example, why does f(x) approach infinity as x approaches infinity and x approaches negative infinity when a polynomial has an even degree with a positive leading coefficient?
Could you explain this to me algebraically and not with calculus reasoning!! Need Help!!!!!

If a polynomial, p(x), has an even degree, then for some n belonging to the positive integers,
p(x) = a_2n x²ⁿ + a_2n-1 x^2n-1 + ... + a₁ x + a₀
where a_2n is not zero. Since we are are also given that the leading coefficient is positive, a_2n > 0. Also, no matter what sign x is, negative or positive, x²ⁿ > 0 if x is not zero. Now we come to the crus of the matter. The leading term, a_2n x²ⁿ, is always positive and eventually dominates the rest of the polynomial and thus as x becomes large enough in magnitude, a_2n x²ⁿ and p(x) have the same sign. Thus as x goes to infinity, p(x) goes to infinity.

If you need to prove that dominance by a_2n x²ⁿ, let b_j = \(\displaystyle \frac{a_j}{a_{2n}}; j = 0, 1, 2, 3, ..., 2n\) and consider
p(x) = a_2n x²ⁿ [1 + \(\displaystyle \frac{b_{2n-1}}{x} + \frac{b_{2n-2}}{x^2} + ... + \frac{b_1}{x^{2n-1}} + \frac{b_0}{x^{2n}}] \)
Each of the terms inside the brackets goes to zero as x goes to infinity, so p(x)/(a_2n x²ⁿ) goes to one as x goes to infinity.

 

[howiejackson14]

If a polynomial, p(x), has an even degree, then for some n belonging to the positive integers,
p(x) = a_2n x²ⁿ + a_2n-1 x^2n-1 + ... + a₁ x + a₀
where a_2n is not zero. Since we are are also given that the leading coefficient is positive, a_2n > 0. Also, no matter what sign x is, negative or positive, x²ⁿ > 0 if x is not zero. Now we come to the crus of the matter. The leading term, a_2n x²ⁿ, is always positive and eventually dominates the rest of the polynomial and thus as x becomes large enough in magnitude, a_2n x²ⁿ and p(x) have the same sign. Thus as x goes to infinity, p(x) goes to infinity.

If you need to prove that dominance by a_2n x²ⁿ, let b_j = \(\displaystyle \frac{a_j}{a_{2n}}; j = 0, 1, 2, 3, ..., 2n\) and consider
p(x) = a_2n x²ⁿ [1 + \(\displaystyle \frac{b_{2n-1}}{x} + \frac{b_{2n-2}}{x^2} + ... + \frac{b_1}{x^{2n-1}} + \frac{b_0}{x^{2n}}] \)
Each of the terms inside the brackets goes to zero as x goes to infinity, so p(x)/(a_2n x²ⁿ) goes to one as x goes to infinity.

I'm still not understanding. Could you elaborate a bit more!

 

[Ishuda]

I'm still not understanding. Could you elaborate a bit more!

What are you not understanding? The dominance of a_2n x²ⁿ or what?