Question

How do you find the end behavior and state the possible number of x intercepts and the value of the y intercept given `y=x^5-1`?

Answer 1

See explanation...

Explanation:

Given:

`f(x) = x^5-1`

As with any polynomial, the end behaviour is dictated by the term of highest degree, which in our example is `x^5`.

Since this is of odd degree, with positive coefficient (`1`), the end behaviour is:

`lim_(x->-oo) f(x) = -oo`

`lim_(x->+oo) f(x) = +oo`

We can find the `y` intercept by evaluating `f(0)`:

`f(0) = 0^5-1 = -1`

So the `y` intercept is at `(0, -1)`

Note that:

`f(1) = 1^5-1 = 0`

So there is an `x` intercept at `(1, 0)`

and `f(x)` has a factor `(x-1)`:

`x^5-1 = (x-1)(x^4+x^3+x^2+x+1)`

The remaining quartic has only complex zeros, which are the non-real fifth roots of 1. It has real quadratic factorisation:

`x^4+x^3+x^2+x+1`

`= (x^2+1/2(sqrt(5)+1)x+1)(x^2+1/2(sqrt(5)-1)x+1)`

The zeros form the vertices of a regular pentagon in the complex plane.

Using de Moivre's formula we can express the complex zeros as:

`cos((2pi)/5)+i sin((2pi)/5)`

`cos((4pi)/5)+i sin((4pi)/5)`

`cos((6pi)/5)+i sin((6pi)/5)`

`cos((8pi)/5)+i sin((8pi)/5)`

So the only `x` intercept is `(1, 0)`.

graph{x^5-1 [-5.52, 4.48, -3.26, 1.74]}