# 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?

Jan 2, 2017

See explanation...

#### Explanation:

Given:

$f \left(x\right) = {x}^{5} - 1$

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:

${\lim}_{x \to - \infty} f \left(x\right) = - \infty$

${\lim}_{x \to + \infty} f \left(x\right) = + \infty$

We can find the $y$ intercept by evaluating $f \left(0\right)$:

$f \left(0\right) = {0}^{5} - 1 = - 1$

So the $y$ intercept is at $\left(0 , - 1\right)$

Note that:

$f \left(1\right) = {1}^{5} - 1 = 0$

So there is an $x$ intercept at $\left(1 , 0\right)$

and $f \left(x\right)$ has a factor $\left(x - 1\right)$:

${x}^{5} - 1 = \left(x - 1\right) \left({x}^{4} + {x}^{3} + {x}^{2} + x + 1\right)$

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

${x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left({x}^{2} + \frac{1}{2} \left(\sqrt{5} + 1\right) x + 1\right) \left({x}^{2} + \frac{1}{2} \left(\sqrt{5} - 1\right) x + 1\right)$

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:

$\cos \left(\frac{2 \pi}{5}\right) + i \sin \left(\frac{2 \pi}{5}\right)$

$\cos \left(\frac{4 \pi}{5}\right) + i \sin \left(\frac{4 \pi}{5}\right)$

$\cos \left(\frac{6 \pi}{5}\right) + i \sin \left(\frac{6 \pi}{5}\right)$

$\cos \left(\frac{8 \pi}{5}\right) + i \sin \left(\frac{8 \pi}{5}\right)$

So the only $x$ intercept is $\left(1 , 0\right)$.

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