To find : which of the functions below could have created this graph.
Solution : We see from the graph both end point are below
By the End Behavior of the function : if the degree of polynomial is even and leading coefficient is negative then the both end point of graph would be below.
We can see from option c , it has degree = 4 (even) and leading coefficient is negative.
It is B, the coefficient of the x-term. When you multiply this out, you get x*x+x*q+x*p+p*q, or x^2+x(p+q)+p*q Notice how the x term is multiplied by p+q.