relative maximum at A(-2, 4)
relative minimum at B(1, 1)
relative maximum at C(5, 7)
| /•\ C
| \•/ B
With no other critical points and 3 turning points the degree is 4 and there are 4 zeros, but only two will be real because the graph will cross the x-axis in only 2 points.
CORRECTION: There could be 4 + 2n, n integer, zeros; but only 2 will be real. I found that using a degree 4 p(x) wouldn't go through all three points. So I changed my p'(x) to p’(x) = a(x+2)(x-1)(x-5)((x-b)^2+c^2) which made the degree of p(x) six, with
2 real zeros and 4 imaginary zeros.
Or, using calculus, p’(x) = a(x+2)(x-1)(x-5). which has degree 3, so the degree of p(x) = 3 + 1 = 4.
4th degree ==> 4 zeros; two are real x-intercepts.
GeoGebra gives the numeric coefficients for a 4th degree polynomial function, p(x), and a 3rd degree polynomial function for the derivative of p(x), p'(x).
p(x) IS DEGREE 4, p'(x) IS DEGREE 3.