First, we begin by finding the derivative of the function h(x) = 6 - 4x³. The derivative of a constant is 0 and the derivative of -4x³ using power rule (dy/dx x^n = n * x^(n-1)) is -12x². Thus, the derivative of the function, also known as the slope of the tangent line to the curve at any point x, is h'(x) = -12x².
Second, we substitute x=1 into the derivative to calculate the slope of the tangent line at the point (1,2). When we do this, we find h'(1) = -12(1)² = -12. So, the slope of the tangent line to the curve at the point (1,2) is -12.
Next, we use point-slope form to find the equation of the tangent line. Point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substituting our point (1,2) and our slope -12 in, we get y - 2 = -12(x - 1). We then simplify this to get y = -12x + 14.
Lastly, we convert our equation into slope-intercept form. Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. Looking at our equation y = -12x + 14, we can see it is already in slope-intercept form. Hence, the equation of the tangent line to the curve y = 6 - 4x³ at the point (1,2) is y = -12x + 14, where the slope m is -12 and the y-intercept b is 14.