Find the quadratic function $y=a x^{2}+b x+c$ whose graph passes through the given points. $(-1,-6),(1,4),(2,12)$
\[
y=\square
\]

Step 1 ：Substitute the coordinates of the points into the equation \(y=ax^2+bx+c\) to get the following system of equations: \(-6=a(-1)^2+b(-1)+c\), \(4=a(1)^2+b(1)+c\), \(12=a(2)^2+b(2)+c\)

Step 2 ：This gives us the system of equations: \(-6=a-b+c\), \(4=a+b+c\), \(12=4a+2b+c\)

Step 3 ：Subtract the first equation from the second to eliminate \(c\): \(4-(-6)=a+b+c-(-a+b-c)\) which simplifies to \(10=2a\)

Step 4 ：Solve for \(a\) to get \(a=5\)

Step 5 ：Substitute \(a=5\) into the first and second equations to get: \(-6=5-b+c\) and \(4=5+b+c\)

Step 6 ：Subtract the first equation from the second to eliminate \(b\): \(4-(-6)=5+b+c-(5-b+c)\) which simplifies to \(10=2b\)

Step 7 ：Solve for \(b\) to get \(b=5\)

Step 8 ：Substitute \(a=5\) and \(b=5\) into the first equation to solve for \(c\): \(-6=5(1)-5(1)+c\) which simplifies to \(-6=-5+c\)

Step 9 ：Solve for \(c\) to get \(c=-1\)

Step 10 ：Therefore, the quadratic function is \(y=5x^2+5x-1\)

Step 11 ：Check if this function passes through the given points: For \((-1,-6)\), \(y=5(-1)^2+5(-1)-1=-6\); For \((1,4)\), \(y=5(1)^2+5(1)-1=4\); For \((2,12)\), \(y=5(2)^2+5(2)-1=12\)

Step 12 ：So, the function \(y=5x^2+5x-1\) does indeed pass through the points \((-1,-6),(1,4),(2,12)\)

Step 13 ：\(\boxed{y=5x^2+5x-1}\) is the final answer