From a trinomial to the sign on the axis
The inequality ax2+bx+c≷0ax^2+bx+c\gtrless0, a≠0a\neq0, can be solved in two consistent ways: 1) Geometrically: the graph y=ax2+bx+cy=ax^2+bx+c is a parabola; find where it lies above or below the xx-axis. 2) Algebraically: factor a(x−x1)(x−x2)a(x-x_1)(x-x_2) when D>0D>0 (or handle a double root when D=0D=0, no real roots when D<0D<0) and apply the sign-chart method to the factors. Both approaches give the same set of xx.
If a>0a>0, branches open up — a "bowl"; if a<0a<0 — down
Roots are xx-intercepts (if any exist)
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