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 $y^{\prime }$ or $f^{\prime }(x)$, gives us the slope of the tangent line to the curve at any point $x$. Using the power rule for differentiation, we get $y^{\prime }=3x^2-6x$.

Step 2 ：Step2: Then, we plug $x=1$ into $y^{\prime }$ to find the slope of the tangent line at the point $(1,0)$. We get $y^{\prime }(1)=3\ast 1^2-6\ast 1=-3$. 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: $y=-3x+3$.