According to the number of zeros and the behavior of the function, it is found that f is of the 3rd degree.
A function of the nth degree has n zeros, that is, n values of x for which f(x) = 0.
- In this problem, there is a zero at [tex]x = -1[/tex].
- Then, at [tex]x = 1[/tex], there is another zero.
- However, if it was a quadratic equation, after [tex]x = 1[/tex], the function would decrease, assuming negative values. Since it increases, we have to assume that this zero at x = 1 has multiplicity 2, that is, it is equivalent to 2 zeros, and thus, the function is of the 3rd degree.
A similar problem is given at https://brainly.com/question/11088875
