14) We have the graph of an odd-degree polynomial.
It has a real root at x = 1 and two imaginary (conjugate) roots.
We can write the polynomial in general terms as:
[tex]p(x)=a(x^2+b^2)(x-c)[/tex]
where x²+b² is the factor that correspond to the imaginary roots and (x-c) is the factor for the real root.
Parameter a is the cubic coefficient.
We know that the real root is x = 1, so c = 1.
We know can look at two known points in order to find a and b.
One point is (0,1) and the other is (-1,2).
Then, we can write for (0,1):
[tex]\begin{gathered} p(0)=1 \\ a(0^2+b^2)(0-1)=1 \\ a\cdot b^2(-1)=1 \\ ab^2=-1 \end{gathered}[/tex]
Now, if we use the point (-1,2), we get:
[tex]\begin{gathered} p(-1)=2 \\ a((-1)^2+b^2)(-1-1)=2 \\ a(1+b^2)(-2)=2 \\ a+ab^2=\frac{2}{-2} \\ a+ab^2=-1 \end{gathered}[/tex]
