Sketch three upper solution relations and qualify their endpoints

$\int y\,dy=\int(2x-1)\,dx$ gives $y^2/2=x^2-x+A$, hence $y^2=2x^2-2x+C$ with C=2A. Since y is nonnegative, use the upper square root where real.

The source enters $Y_1=\sqrt{2x^2-2x}$ (C=0), $Y_2=\sqrt{2x^2-2x-1}$ (C=−1), $Y_3=\sqrt{2x^2-2x+1}$ (C=1). For C=−1 the geometric x-intercepts are $(1\pm\sqrt3)/2\approx-0.366,1.37$. For C=0 they are 0 and 1. The C=1 curve is smooth for all real x, with minimum $(1/2,1/\sqrt2)\approx(0.5,0.707)$ and y-intercept 1.

For C=−1 and C=0, the zero-height endpoints are valid points of the implicit squared relation, but not classical differentiable solutions of the original equation there: at y=0, the left side would be 0 while 2x−1 is nonzero. The classical solution branches therefore use the open intervals beyond those endpoints; the displayed figure marks them open while retaining their limiting coordinates. The Teacher note labels family sketches as outside the stated examination syllabus; this exercise is retained as source enrichment rather than removed.