The present problem presents the following equation:
[tex]F(x)=x^2(x+1)^2(x-1)[/tex]It follows the general form of a polynomial that which follows:
[tex]F_{\text{general}}(x)=(x-a)^{na}(x-b)^{nb}(b-c)^{nc}\ldots[/tex]Where a, b, c... stands for the zeros of the function, and na, nb, nc... for its multiplicity.
According to the figure and the rule is given, the F presents three zeros:
[tex]\begin{gathered} P1\colon(-1,0) \\ P2\colon(0,0) \\ P3\colon(1,0) \end{gathered}[/tex]Analyzing each of them we are able to say that, P1 AND P2 have multiplicity equal to 2, and the graph touches but does not crosses the x-axis, while it crosses at P3.
The y-intercept is the point where x = 0, which implies P2.
From the solution presented above, we are able to answer the problem as follows:
