[proofplan] The equation defining the affine real locus is solved pointwise in the vertical coordinate: for a fixed real abscissa $x$, a real ordinate $y$ exists exactly when $f(x) \geq 0$, and the ordinates are precisely $\pm \sqrt{f(x)}$. It remains to determine the sign set of the monic cubic $f$. In the one-real-root case, factor $f$ as a real linear factor times a positive quadratic; in the three-real-root case, factor $f$ into three ordered linear factors and read off the sign on the intervals cut out by the roots. The no-repeated-root hypothesis rules out tangencies at roots and hence leaves only these two real-root patterns. [/proofplan]

[step:Identify the real affine locus with the two square-root graphs]Let $E(\mathbb{R})$ denote the real affine locus \begin{align*} E(\mathbb{R}) := \{(x,y) \in \mathbb{R}^2 : y^2=x^3+ax+b\}. \end{align*} Define the nonnegative set \begin{align*} S := \{x \in \mathbb{R} : f(x) \geq 0\}. \end{align*} Let $(x,y) \in E(\mathbb{R})$. By definition of $E(\mathbb{R})$, the coordinates satisfy $y^2 = f(x)$. Since $y^2 \geq 0$ for every $y \in \mathbb{R}$, we have $x \in S$. The real square root function $\sqrt{\cdot}: [0,\infty) \to [0,\infty)$ is defined by the condition that $\sqrt{t}$ is the unique nonnegative real number whose square is $t$. Therefore $y^2 = f(x)$ implies \begin{align*} y = \sqrt{f(x)} \quad\text{or}\quad y = -\sqrt{f(x)}. \end{align*} Thus \begin{align*} E(\mathbb{R}) \subset \{(x,\sqrt{f(x)}) : x \in S\} \cup \{(x,-\sqrt{f(x)}) : x \in S\}. \end{align*} Conversely, if $x \in S$, then $f(x) \in [0,\infty)$, so both $\sqrt{f(x)}$ and $-\sqrt{f(x)}$ are [real numbers](/page/Real%20Numbers) and satisfy \begin{align*} (\sqrt{f(x)})^2 = f(x), \qquad (-\sqrt{f(x)})^2 = f(x). \end{align*} Hence both corresponding points belong to $E(\mathbb{R})$. This proves \begin{align*} E(\mathbb{R}) = \{(x,\sqrt{f(x)}) : x \in S\} \cup \{(x,-\sqrt{f(x)}) : x \in S\}. \end{align*}[/step]

[guided]Let $E(\mathbb{R})$ denote the real affine locus \begin{align*} E(\mathbb{R}) := \{(x,y) \in \mathbb{R}^2 : y^2=x^3+ax+b\}. \end{align*} Fix a real number $x \in \mathbb{R}$. The equation of the curve becomes a one-variable equation in the ordinate $y$: \begin{align*} y^2 = f(x). \end{align*} A real square is always nonnegative, so if there exists $y \in \mathbb{R}$ with $y^2 = f(x)$, then necessarily $f(x) \geq 0$. This proves that every real point of the curve has its first coordinate in \begin{align*} S := \{x \in \mathbb{R} : f(x) \geq 0\}. \end{align*} Now suppose $x \in S$. Then $f(x)$ lies in the domain $[0,\infty)$ of the real square root function $\sqrt{\cdot}: [0,\infty) \to [0,\infty)$, where $\sqrt{t}$ denotes the unique nonnegative real number whose square is $t$. Therefore the real solutions of $y^2 = f(x)$ are exactly \begin{align*} y = \sqrt{f(x)} \quad\text{and}\quad y = -\sqrt{f(x)}. \end{align*} If $f(x) = 0$, these two displayed values coincide, giving the single ordinate $y=0$; if $f(x)>0$, they are two distinct ordinates. Combining the necessary condition $x \in S$ with the explicit description of the two possible ordinates gives \begin{align*} E(\mathbb{R}) = \{(x,\sqrt{f(x)}) : x \in S\} \cup \{(x,-\sqrt{f(x)}) : x \in S\}. \end{align*} Thus the geometry of the real affine locus is completely controlled by the sign set of the cubic $f$.[/guided]

[step:Determine the sign set when the cubic has one real root]Assume that $f$ has exactly one real root, denoted by $r \in \mathbb{R}$. Since $f$ is monic and $r$ is a root, polynomial division gives a monic quadratic polynomial $q: \mathbb{R} \to \mathbb{R}$ defined by \begin{align*} q(x) = x^2 + rx + r^2 + a \end{align*} such that \begin{align*} f(x) = (x-r)q(x) \end{align*} for every $x \in \mathbb{R}$. The polynomial $q$ has no real root. Indeed, if $s \in \mathbb{R}$ satisfied $q(s)=0$, then $f(s)=(s-r)q(s)=0$. If $s \neq r$, this would give a second real root of $f$, contradicting the assumption that $r$ is the only real root. If $s=r$, then $q(r)=0$ and the factorization $f(x)=(x-r)q(x)$ would make $r$ a repeated root of $f$, contradicting nonsingularity. Since $q$ is monic, its leading coefficient is positive. A real quadratic with positive leading coefficient tends to $+\infty$ as $|x| \to \infty$; if it took a nonpositive value, continuity and the intermediate value property for polynomials would force a real zero. Because $q$ has no real zero, it follows that \begin{align*} q(x) > 0 \end{align*} for every $x \in \mathbb{R}$. Therefore the sign of $f(x) = (x-r)q(x)$ is the sign of $x-r$. More explicitly, if $x<r$, then $f(x)<0$. At $x=r$, one has $f(r)=0$. If $x>r$, then $f(x)>0$. It follows that \begin{align*} S = \{x \in \mathbb{R} : f(x) \geq 0\} = [r,\infty). \end{align*}[/step]

[guided]Assume that $f$ has exactly one real root, and denote that root by $r \in \mathbb{R}$. Because $r$ is a root, $f(r)=r^3+ar+b=0$, so $b=-r^3-ar$. Define the monic quadratic factor $q: \mathbb{R} \to \mathbb{R}$ by \begin{align*} q(x) = x^2 + rx + r^2 + a. \end{align*} Then direct multiplication gives \begin{align*} (x-r)q(x)=(x-r)(x^2+rx+r^2+a)=x^3+ax-r(r^2+a)=x^3+ax+b=f(x) \end{align*} for every $x \in \mathbb{R}$. The key point is that the quadratic factor cannot change sign, because it has no real zero. We verify this. If there were $s \in \mathbb{R}$ with $q(s)=0$, then \begin{align*} f(s) = (s-r)q(s)=0. \end{align*} If $s \neq r$, this would be a second real root of $f$, contradicting the assumption that $r$ is the only real root. If $s=r$, then the factorization $f(x)=(x-r)q(x)$ together with $q(r)=0$ would make $r$ a repeated root of $f$, contradicting the nonsingularity hypothesis as verified in the repeated-root exclusion step. Hence $q$ has no real root. Since $q$ is monic, its leading coefficient is positive, and therefore $q(x) \to +\infty$ as $|x| \to \infty$. If $q$ took a negative value at some point, the intermediate value property for the polynomial $q$ would give a real zero on each side of that point for sufficiently large $|x|$. If $q$ took the value $0$, it would already have a real zero. Because we proved that $q$ has no real zero, the only remaining possibility is \begin{align*} q(x)>0 \end{align*} for every $x \in \mathbb{R}$. Thus the factor $q(x)$ never affects the sign of $f(x)=(x-r)q(x)$; the sign is exactly the sign of $x-r$. If $x<r$, then $x-r<0$ and $q(x)>0$, so $f(x)<0$. If $x=r$, then $f(r)=0$. If $x>r$, then $x-r>0$ and $q(x)>0$, so $f(x)>0$. Consequently \begin{align*} \{x \in \mathbb{R}: f(x)\geq 0\}=[r,\infty). \end{align*}[/guided]

[step:Determine the sign set when the cubic has three ordered real roots]Assume that $f$ has three real roots $r_1,r_2,r_3 \in \mathbb{R}$ with \begin{align*} r_1 < r_2 < r_3. \end{align*} Because $f$ is monic and has these three roots, its factorization over $\mathbb{R}$ is \begin{align*} f(x) = (x-r_1)(x-r_2)(x-r_3) \end{align*} for every $x \in \mathbb{R}$. The ordered roots divide $\mathbb{R}$ into the four intervals $(-\infty,r_1)$, $(r_1,r_2)$, $(r_2,r_3)$, and $(r_3,\infty)$. On these intervals, the signs of the factors are as follows: If $x<r_1$, then $(x-r_1)<0$, $(x-r_2)<0$, and $(x-r_3)<0$. If $r_1<x<r_2$, then $(x-r_1)>0$, $(x-r_2)<0$, and $(x-r_3)<0$. If $r_2<x<r_3$, then $(x-r_1)>0$, $(x-r_2)>0$, and $(x-r_3)<0$. If $r_3<x$, then $(x-r_1)>0$, $(x-r_2)>0$, and $(x-r_3)>0$. Multiplying the three signs gives Thus $f(x)<0$ on $(-\infty,r_1)$, $f(x)>0$ on $(r_1,r_2)$, $f(x)<0$ on $(r_2,r_3)$, and $f(x)>0$ on $(r_3,\infty)$. At each root $r_i$, the factorization gives $f(r_i)=0$. Therefore \begin{align*} S = \{x \in \mathbb{R} : f(x) \geq 0\} = [r_1,r_2] \cup [r_3,\infty). \end{align*}[/step]

[guided]The point of ordering the roots is that each linear factor has a fixed sign on each interval cut out by the roots. Since $f$ is monic and has precisely the roots $r_1,r_2,r_3$, we may write \begin{align*} f(x) = (x-r_1)(x-r_2)(x-r_3) \end{align*} for every $x \in \mathbb{R}$. Now examine the intervals determined by $r_1<r_2<r_3$. If $x<r_1$, then $x$ is smaller than all three roots, so all three factors are negative: \begin{align*} (x-r_1)<0,\qquad (x-r_2)<0,\qquad (x-r_3)<0. \end{align*} The product of three negative numbers is negative, so $f(x)<0$ on $(-\infty,r_1)$. If $r_1<x<r_2$, then $x-r_1$ is positive while $x-r_2$ and $x-r_3$ are negative: \begin{align*} (x-r_1)>0,\qquad (x-r_2)<0,\qquad (x-r_3)<0. \end{align*} The product has two negative factors, hence is positive. Thus $f(x)>0$ on $(r_1,r_2)$. If $r_2<x<r_3$, then the first two factors are positive and the last factor is negative: \begin{align*} (x-r_1)>0,\qquad (x-r_2)>0,\qquad (x-r_3)<0. \end{align*} The product is negative, so $f(x)<0$ on $(r_2,r_3)$. Finally, if $x>r_3$, then all three factors are positive: \begin{align*} (x-r_1)>0,\qquad (x-r_2)>0,\qquad (x-r_3)>0, \end{align*} and therefore $f(x)>0$ on $(r_3,\infty)$. At the roots themselves, one of the factors is zero, so \begin{align*} f(r_1)=f(r_2)=f(r_3)=0. \end{align*} Combining the intervals where $f$ is positive with the roots where $f$ is zero gives exactly \begin{align*} \{x \in \mathbb{R}: f(x)\geq 0\} = [r_1,r_2]\cup [r_3,\infty). \end{align*}[/guided]

[step:Use nonsingularity to rule out repeated-root alternatives]We briefly verify the repeated-root exclusion from nonsingularity in this short Weierstrass form. Define the polynomial map $F: \mathbb{R}^2 \to \mathbb{R}$ by \begin{align*} F(x,y)=y^2-x^3-ax-b. \end{align*} Define the derivative polynomial $f': \mathbb{R} \to \mathbb{R}$ by \begin{align*} f'(x)=3x^2+a. \end{align*} The affine cubic is cut out by $F(x,y)=0$, and its partial derivatives are \begin{align*} \partial_x F(x,y)=-3x^2-a, \qquad \partial_y F(x,y)=2y. \end{align*} If $r \in \mathbb{R}$ were a repeated root of $f$, then $f(r)=0$ and $f'(r)=3r^2+a=0$. The point $(r,0)$ would lie on $E(\mathbb{R})$, and both partial derivatives of $F$ would vanish there: \begin{align*} \partial_x F(r,0)=0, \qquad \partial_y F(r,0)=0. \end{align*} This would make $(r,0)$ a singular point, contradicting the nonsingularity hypothesis. Hence no real root of $f$ is repeated. Since a real cubic has either one real root or three real roots counted without repetition after excluding repeated roots, the two cases above exhaust the possible real-root configurations under the stated alternatives. Combining the square-root graph description with the computed sign sets gives the asserted description of $E(\mathbb{R})$ in both cases.[/step]

[guided]We must justify that the nonsingularity assumption really excludes repeated roots of the cubic polynomial. Define the polynomial map $F: \mathbb{R}^2 \to \mathbb{R}$ by \begin{align*} F(x,y)=y^2-x^3-ax-b. \end{align*} The affine real curve is the zero set of $F$, because $F(x,y)=0$ is exactly the equation $y^2=x^3+ax+b$. Also define the derivative polynomial $f': \mathbb{R} \to \mathbb{R}$ by \begin{align*} f'(x)=3x^2+a. \end{align*} The partial derivatives of $F$ are \begin{align*} \partial_x F(x,y)=-3x^2-a, \qquad \partial_y F(x,y)=2y. \end{align*} Suppose, for contradiction, that $r \in \mathbb{R}$ is a repeated root of $f$. In this cubic setting, repeated root means that there is a polynomial $h: \mathbb{R} \to \mathbb{R}$ with \begin{align*} f(x)=(x-r)^2 h(x) \end{align*} for every $x \in \mathbb{R}$. Substituting $x=r$ gives $f(r)=0$, and differentiating the displayed identity gives \begin{align*} f'(x)=2(x-r)h(x)+(x-r)^2h'(x), \end{align*} so substituting $x=r$ gives $f'(r)=0$. Thus \begin{align*} f(r)=0, \qquad f'(r)=0. \end{align*} Since $f'(r)=3r^2+a$, the second equality gives $3r^2+a=0$. The first equality gives \begin{align*} 0=f(r)=r^3+ar+b, \end{align*} so the point $(r,0)$ lies on the affine cubic. At this point the two partial derivatives of $F$ vanish: \begin{align*} \partial_x F(r,0)= -3r^2-a = 0, \qquad \partial_y F(r,0)=0. \end{align*} Thus $(r,0)$ is a singular point of the affine plane curve cut out by $F=0$, contradicting the hypothesis that the short Weierstrass cubic is nonsingular. Therefore no real root of $f$ is repeated. A real cubic has at least one real root, and after repeated roots are excluded its real-root configurations are exactly one real root or three distinct real roots. These are precisely the two sign analyses already carried out.[/guided]