How do you graph #f(x)=x^5-2# using zeros and end behavior?

1 Answer
Dec 4, 2016

See explanation.


The number of changes in the signs of the coefficients in

f(x) and f(-x) are 1 and 0, respectively.

As the degree of f is odd, we conclude that there is no negative root

and there is exactly one positive root. The other four zeros are

complex, occurring in two conjugate pairs.

Easily, the positive root is 2^(1/5)=1.1487, nearly.

As # x to +-oo, y to +-oo#.

The y-intercept ( x = 0 ) is -2.

The x-intercept ( y = 0 ) is 1.1487, nearly.

All this behavior is well illustrated by the inserted graph

There are no asymptotes

graph{x^5-2 [-20, 20, -10, 10]}