# will a polynomial function with a leading coefficient of 13 and degree 8 have a range of all real numbers? ?

### 6 Answers

- PinkgreenLv 71 month ago
Let y=13x^8+a1*x^7+a2*x^6+...+a8, where ai, 1=<i<=8 are constants.

=>

y=x^8(13+a1/x+a1/x^2+...+a8/x^8)

=>

limit y = +infinity

x->+/-inf.

=>

the range=[0, +inf.), if a8=0

or

the range=(0, +inf.), if a8=/=0

- PuzzlingLv 71 month ago
A polynomial with an even degree will have the same starting and ending behavior. In the case of a positive leading coefficient (e.g. 13), that means the left end and the right end are both going up to infinity.

They will never go down to negative infinity (there must be a minimum) and hence it can't have a *range* that is all real numbers.

- MorningfoxLv 71 month ago
A poly like that starts with 13x^8. For very large values of x, that term dominates all the others. So for very large values of x, we have 13x^8, as x goes to infinity. And in the limit, that is positive infinity. Some for x goes to negative infinity .... the function goes to positive infinity.

The function does not go to negative infinity, so it has a finite lower value. The range does not include numbers less than that lower value. The function does NOT have a range of all real numbers.

- MathLv 71 month ago
Real numbers have no numerical range.

They're infinite. I don't think any such polynomial can exist whose graph covers the entire y axis. It will have a min and max.

I think you mean a domain of all reals. Please check

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- davidLv 71 month ago
No. --- Even degree must have an absolute max or an absolute min (not both) ... this limits the range of the function

==== Think of a quadratic ,,, y = x^2 + 9x + 6 ... has an abs min --