Final answer:
By setting the equation of the line equal to the equation of the curve and solving for the points of intersection, we find that the least possible value of m such that the quadratic has two distinct real roots is m = -1.
Step-by-step explanation:
To find the least possible value of m where the line y = mx intersects the curve y = 1/4(x^3 - 6x^2 + 8x) at three points, we need to set the equation of the line equal to the equation of the curve and solve for x to find the points of intersection.
The intersection points are where mx = 1/4(x^3 - 6x^2 + 8x), which simplifies to 0 = x^3 - 6x^2 + 8x - 4mx. The factor x can be factored out, leading to x(x^2 - 6x + 8 - 4m) = 0, which indicates that one of the intersections is at x=0.
To find the least value of m such that the quadratic x^2 - 6x + 8 - 4m has two distinct real roots, we can use the discriminant D = b^2 - 4ac. For the quadratic to have two distinct real roots, the discriminant must be greater than zero. Thus, (-6)^2 - 4(1)(8 - 4m) > 0, which simplifies to 36 - 32 + 16m > 0 and 16m > -4, so m > -1. The least possible value of m that meets this condition is m = -1.