We are given the quadratic function c(p) = p^2 - 28p + 250 and we need to rewrite this in such a way that the minimum value is easy to find.
Remember that the vertex
[tex]f(x)=a(x-h)^2+k[/tex]
where (h, k) is the vertex or the minimum point.
We use completing the squares method to do this. We divide the second term, -28p, by 2p, then square it.
[tex](\frac{-28p}{2p})^2=(-14)^2=196[/tex]
We add 196 to p^2 - 28p to make it equal to (p - 14)^2, but since it will change the value of the equation, we need to subtract the same value from 250 so that the net effect is zero.
[tex]\begin{gathered} c(p)=p^2-28p+250 \\ c(p)=(p^2-28p+196)+250-196 \\ c(p)=(p-14)^2+54 \end{gathered}[/tex]
The equation is c(p) = (p - 14)^2 + 54.
