Find the equation of the tangent line to the curve \(y = x^3 - 3x^2 + 2\) at the point \((1,0)\)

Step 1 ：Step1: First, we find the derivative of \(y = x^3 - 3x^2 + 2\). The derivative of the function, denoted as \(y'\) or \(f'(x)\), gives us the slope of the tangent line to the curve at any point \(x\). Using the power rule for differentiation, we get \(y' = 3x^2 - 6x\).

Step 2 ：Step2: Then, we plug \(x = 1\) into \(y'\) to find the slope of the tangent line at the point \((1,0)\). We get \(y'(1) = 3*1^2 - 6*1 = -3\). So, the slope of the tangent line is -3.

Step 3 ：Step3: Now that we have the slope of the tangent line and the point \((1,0)\), we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form of a line is \(y - y1 = m(x - x1)\), where \((x1, y1)\) is a point on the line and \(m\) is the slope of the line. Plugging in the values we have, we get \(y - 0 = -3(x - 1)\). Simplifying this gives us the equation of the tangent line: \(y = -3x + 3\).