2 answers
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y' = 4x
so you want a point (h,2h^2+3) on the curve such that
((2h^2+3)-3)/(h-2) = 4h
h = 0 or 4
So the points on the curve are (0,3) and (4,35)
That makes the lines
y=3
y = 16x -29
See the graphs at
www.wolframalpha.com/input?i=plot+y+%3D+2x%5E2+%2B+3%2C+y%3D3%2C+y%3D16x-29+for+-2+%3C%3D+x+%3C%3D+6
Do the other in like wise.