f(x)=ax²+bx+c, f'(x)=2ax+b.
At P(0,0), f'(0)=b and at Q(16,y), f'(16)=32a+b.
L1 is tangential at P, and its gradient has to be that of the parabola, so b=0.8 and f'(16)=32a+0.8.
For L1, y=0.8x+k. The point P(0,0) lies on L1 and f(x), therefore:
k=c=0,
f(x)=ax²+0.8x,
y=0.8x is L1.
At Q(16,f(16)), f'(16)=32a+0.8=-3.3 because L2 has the same gradient as f(x) at Q, so 32a=-4.1, a=-4.1/32. y=f(16)=256a+12.8=-32.8+12.8=-20. Therefore Q is at (16,-20).
f(x)=-4.1x²/32+0.8x.
For L2, y=-3.3x+k, so plug in x=16, y=-20:
-20=-52.8+k, k=32.8 and L2 is y=-3.3x+32.8.
The graph shows two lines and a parabola where L1 is red, L2 is green and f(x) is blue. These are shown without segmentation.
L1 ceases at P where it joins f(x). The next segment tracks f(x) up to Q, where it joins the start of L2, and segment L2 continues downwards.