y = 5x^2 +8x ----------------(1)
So, tangent to the curve is y' = 10x +8
Also, the slope of the tangent to the curve that is parallel to 12x+y=4 or y=-12x+4 is -12
Therefore,
10x + 8 = -12
or 10x =-20
or x=-2
putting the value of x in (1) we get:
y = 5(4) +8(-2) = 4
So the equation of L is y-4=-12(x+2)
or y = -12x -24 +4
or y = -12x -20
So the y intercept of line L is -20