Find the equation of the tangent line to the graph of $f(x)$ at the $(x, y)$-coordinate indicated below.
\[
f(x)=\left(4 x^{2}-3 x-3\right)\left(-x^{2}+3\right) ;(1,-4)
\]

Step 1 ：Given the function \(f(x)=(4x^{2}-3x-3)(-x^{2}+3)\) and the point \((1,-4)\).

Step 2 ：Find the derivative of the function \(f(x)\).

Step 3 ：\(f'(x) = -2x(4x^{2} - 3x - 3) + (3 - x^{2})(8x - 3)\).

Step 4 ：Substitute \(x = 1\) into \(f'(x)\) to find the slope of the tangent line at that point.

Step 5 ：The slope of the tangent line at \(x = 1\) is \(14\).

Step 6 ：Use the point-slope form of the equation of a line, \(y - y_{1} = m(x - x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) is the given point.

Step 7 ：Substitute \(m = 14\), \(x_{1} = 1\), and \(y_{1} = -4\) into the equation to find the equation of the tangent line.

Step 8 ：The equation of the tangent line is \(y = 14x - 10\).

Step 9 ：Final Answer: The equation of the tangent line to the graph of \(f(x)\) at the point \((1,-4)\) is \(\boxed{y = 14x - 10}\).