Find $\mathrm{b}$ and $\mathrm{c}$ so that $y=14 x^{2}+b x+c$ has vertex $(-1,4)$ \[ b= \] \[ c= \]

Step 1 ：The vertex form of a parabola is given by \(y=a(x-h)^{2}+k\) where \((h,k)\) is the vertex of the parabola.

Step 2 ：In this case, we know that \(a=14\), \(h=-1\) and \(k=4\).

Step 3 ：We can expand this equation to get it in the form \(y=ax^{2}+bx+c\).

Step 4 ：Doing so, we get \(y = 14*(x + 1)^{2} + 4\).

Step 5 ：Expanding this equation, we get \(y = 14*x^{2} + 28*x + 18\).

Step 6 ：Comparing coefficients, we find that \(b = 28\) and \(c = 18\).

Step 7 ：Final Answer: The coefficients \(b\) and \(c\) are \(\boxed{28}\) and \(\boxed{18}\) respectively.