x^2+4x+3...is (x+1)*(x+3)...parabola that go up
y=0 at x=-1 or x=-3
kompare tu 0.5*x^2+4x+3...y=0 at x=-0.837 or x=-7.16, so yu shift kerv left
(1/2) x^2 gotta be less than 1*x^2, so parabola go up less steep
extreem...0*x^2 become y=4x+3...strate line