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#?

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Answer 1 · 2017-01-02T12:21:32

See 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]}