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 \(\boxed{y = 14x - 10}\).