# explain in terms of graphs why a polynomial of odd degree must have at least one real zero

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### 3 Answers

A polynomial is a function of the type `a_nx^n+a_(n-1)x^(n-1)+...+a_1x+a_0` with `a_iinRR` and `ninNN` . We can examine the end-behavior of these functions:

Suppose n is odd. We take ` `` `` `the limit as `n->+-oo` of `f(x)=a_nx^n+...a_0` .

Suppose `a_n>0` . Then `lim_(n->oo)f(x)=oo` and `lim_(n->-oo)f(x)=-oo` . The function is continuous everywhere, so by the intermediate value theorem there is at least one x such that f(x)=0. An x-intercept is a real zero, so the function has at least one zero.

Suppose `a_n<0` . Then `lim_(n->oo)f(x)=-oo` and `lim_(n->oo)f(x)=oo` . Again since the function is continuous the intermediate value theorem applies and there exists at least one real zero.

The graph of an odd-degree function has the following properties:

If the leading coefficient is positive then the end-behavior is that the graph goes to negative infinity as x goes to neg. infinity, and the graph increases without bound as x increases, so the graph must pass through the x-axis implying a real root.

An analogous argument holds for negative leading coefficient.

You are correct. The limit should be as x approaches infinity, etc...

If the leading coefficient is positive then the term `a_nx^n` goes to negative infinity as x goes to negative infinity; `a_n>0,x^n<0` . (The term `a_nx^n` is the only term we are concerned with since it is far larger in absolute value than the other terms combined for sufficiently large/small x.) As x goes to positive infinity, `a_nx^n` goes to positive infinity since `a_n>0,x^n>0` .

If the leading coefficient is negative (the red graph above) then the term `a_nx^n` goes to positive infinity as x decreases without bound; `a_n<0,x^n<0` . Also `a_nx^n` goes to negative infinity as x grows without bound since `a_n<0,x^n>0` .

I am a little confused. If you are assuming that n is a natural number, how are you studying the behavior of the function as n approaches negative infinity? Also how did you conclude that the function approaches neg inf as n tends to neg inf?