Recast a quartic as a rational-curve intersection

$y=x/2+7/4+11/[4(2x-1)]$. Therefore asymptotes are x=1/2 and y=x/2+7/4. Intercepts are $((-3\pm\sqrt5)/2,0)$, approximately (−2.62,0),(−0.382,0), and (0,−1). Stationary points are $((1-\sqrt{11})/2,2-\sqrt{11}/2)$, maximum≈(−1.16,0.342), and $((1+\sqrt{11})/2,2+\sqrt{11}/2)$, minimum≈(2.16,3.66). All source rounded labels are retained.

$2x^4+6x^3+2x^2+2x-1=0$ gives $2x^2(x^2+3x+1)=-(2x-1)$, hence $(x^2+3x+1)/(2x-1)=-1/(2x^2)$. Neither excluded input 0 nor 1/2 is a root of the original polynomial.

Add $y=-1/(2x^2)$ to the source rational graph. It has a vertical asymptote x=0, horizontal asymptote y=0 and both branches below the x-axis. It meets C at two points, so the quartic has exactly two distinct real roots, as the source concludes.