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

To determine the slope of the tangent line to the curve at a specific point, we can find the derivative of the function and evaluate it at that point.

Step 1: Find the derivative (dy/dx) of the given function y = 2x^3 - 3x^2 + 1.
To find the derivative, we differentiate each term with respect to x using the power rule for differentiation. For a term of the form ax^n, the derivative is dx/dx = anx^(n-1).

Taking the derivative of each term, we get:
dy/dx = d/dx (2x^3) - d/dx (3x^2) + d/dx (1)
= 6x^2 - 6x

Step 2: Evaluate the derivative at the given point (1,0).
Plug the x-coordinate of the given point (1) into the derivative expression:
dy/dx = 6(1)^2 - 6(1)
= 6 - 6
= 0

The slope of the tangent line is equal to the value of the derivative at the point (1,0), which is 0. Therefore, the slope of the tangent line to the curve y = 2x^3 - 3x^2 + 1 at the point (1,0) is 0.