how is graph of y=x^2+4x+3 affectd when coefficient of x^2 changed to smaller positive number?

Question

i need to explain the answer

Answer 1

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