Find $b$ 'and $c$ so that $y=14 x^{2}+b x+c$ has vertex $(-9,-9)$.
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b=
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\[
c=
\]
Answer
Final Answer: The values of \(b\) and \(c\) that make the equation \(y=14 x^{2}+b x+c\) have vertex \((-9,-9)\) are \(b=\boxed{252}\) and \(c=\boxed{1125}\).
Step 1 ：We are given the vertex form of a parabola as \(y=a(x-h)^{2}+k\) where \((h,k)\) is the vertex of the parabola.
Step 2 ：In this case, we know that \(a=14\), \(h=-9\), and \(k=-9\).
Step 3 ：We can expand this equation to get it in the form \(y=ax^{2}+bx+c\).
Step 4 ：After expanding, we get the equation as \(y=14x^{2} + 252x + 1125\).
Step 5 ：By comparing coefficients, we find that \(b=252\) and \(c=1125\).
Step 6 ：Final Answer: The values of \(b\) and \(c\) that make the equation \(y=14 x^{2}+b x+c\) have vertex \((-9,-9)\) are \(b=\boxed{252}\) and \(c=\boxed{1125}\).
