First, solve f(x)=4x-3x^2=0, or x(4-3x)=0 => x=0, x=4/3 The area enclosed by the parabola over the x-axis is therefore A=integral f(x)dx from 0 to 4/3=[2x^2-x^3] from 0 to 4/3 = 32/27 Let the line intersect the parabola at a point (a,f(a)) such that the area bounded by the line, the parabola and the x-axis is half of A, or A/2, then the area consists of a triangle and a section below the parabola, the area is therefore a*f(a)/2 + integral f(x)dx from a to 4/3 = A/2 = 16/27 => 2a^2-3a^3/2+a^3-2a^2+32/27=16/27 => (1/2)a^3=16/27 a=(32/27)^(1/3) =(2/3)(4^(1/3)) =1.058267368...
Slope of line is therefore m=y/x=f(a)/a=4-2(4^(1/3)) =0.825197896... (approx.)
