(1 point) For what values of $x$ is the tangent line of the graph of \[ f(x)=10 x^{3}+15 x^{2}-362 x+120 \] parallel to the line $y=1.9-2 x$ ? (If there are multiple values then separate them with commas.) \[ x= \] Preview My Answers Submit Answers You have attempted this problem 1 time.

Step 1 ：The slope of the line \(y=1.9-2x\) is -2. The tangent line to the function \(f(x)\) will be parallel to this line when the derivative of \(f(x)\) is equal to -2.

Step 2 ：We need to find the derivative of \(f(x)\), set it equal to -2, and solve for \(x\).

Step 3 ：The derivative of \(f(x) = 10x^{3} + 15x^{2} - 362x + 120\) is \(f'(x) = 30x^{2} + 30x - 362\).

Step 4 ：Setting \(f'(x)\) equal to -2 gives us the equation \(30x^{2} + 30x - 362 = -2\).

Step 5 ：Solving this equation gives us the solutions \(x = -4, 3\).

Step 6 ：Final Answer: The values of \(x\) for which the tangent line of the graph of \(f(x)=10 x^{3}+15 x^{2}-362 x+120\) is parallel to the line \(y=1.9-2 x\) are \(\boxed{-4, 3}\).