what are the formulas for the x and y coordinates of the vertex of a general parabola in terms of the coefficients of the general cartesian form? General cartesian form? General parabola: ax^2 + bxy + cy^2 + dx + ey + f = 0 ? Or general cartesian form maybe y = ax^2 + bx + c? No that's standard. General parabola in terms of coefficients of general cartesian form: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, with parabola condition B^2 -4AC =0, and A+C !=0? Actually for parabola, B^2 -4AC =0. Vertex formula? Need to find vertex of general parabola (rotated). That is more involved. If general form is ax^2 + bxy + cy^2 + dx + ey + f =0. Vertex (x0,y0) can be found via completing square after rotation to eliminate xy term. But perhaps user means general parabola y = ax^2 + bx + c? That is general cartesian form? Could be ax^2 + bx + c =0? Hmm. Alternative interpretation: general parabola: y = ax^2 + bx + c, vertex at (-b/(2a), c - b^2/(4a)). Or x = -b/(2a), y = f(x). If it's general conic: Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0. Vertex formula is messy. Maybe they want formula for vertex of parabola in standard form: (x-h)^2 =4p(y-k). But coefficients? Let's assume general form: Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0 with B^2 -4AC=0. We can find vertex via solving gradient =0? Not exactly. Better approach: For parabola, can find axis direction. If B^2 -4AC=0, then either A=0 or C=0 (up to rotation). Without loss, assume A !=0, C=0? Actually B^2=4AC. General case: rotate axes to eliminate xy. Let angle θ such that cot 2θ = (A - C)/B. Then in rotated coordinates (x', y'), equation becomes A' x'^2 + C' y'^2 + D' x' + E' y' + F =0 where A' C' =0. Then vertex is at... We can derive formula. Simpler: If parabola is not rotated, i.e., B=0, then form is A x^2 + C y^2 + D x + E y + F =0. With parabola, A*C=0. Case 1: C=0, A ≠0: A x^2 + D x + E y + F =0 => y = -(A x^2 + D x + F)/E . Vertex x0 = -D/(2A), y0 = E? Wait compute. Actually y = -(A/E) x^2 - (D/E) x - F/E. Vertex x0 = -(-D/E)/(2 * -A/E)= -D/(2A)? Let's compute: a = -A/E, b = -D/E. x0 = -b/(2a)= -(-D/E)/(2 * -A/E)= (D/E)/(-2A/E)= -D/(2A). Yes. y0 = y(x0)= -(A x0^2 + D x0 + F)/E. Similarly for A=0. But general rotated case is messy. Probably the user expects simple formula: x_v = -b/(2a), y_v = -Δ/(4a) ??? Wait for general form y = a x^2 + b x + c, vertex y = c - b^2/(4a). If they ask for general cartesian form ax^2 + by + c x + d =0? Hmm. Alternative general cartesian form: y = ax^2 + bx + c. Maybe they want in terms of coefficients A,B,C of Ax^2 + Bx + C =0? That's quadratic. Better give both: non-rotated and rotated. Given ambiguity, provide general solution. Let's assume general parabola: Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0 with B^2 -4AC=0. We can find vertex via solving linear system derived from completing square. Method: Find axis direction vector v. For parabola, matrix [[A, B/2],[B/2, C]] has rank 1. Let vector u be null eigenvector of matrix: A u_x + (B/2) u_y =0? Actually... Simpler to provide formula for non-rotated. I think most textbooks: general parabola y = a x^2 + b x + c => vertex (-b/(2a), 4ac - b^2 /4a). If they want general form: ax^2 + bx + c =0? No. Maybe they want: For parabola Ax^2 + Bx + C =0? No. Could be they want formula for vertex of parabola in general form: y = a x^2 + b x + c. Provide derivation. Given the instruction to use tools, maybe we need to compute something? No. Possibly they want x = -b/(2a), y = (4ac - b^2)/(4a). Or using discriminant. Let's provide general rotated formula. We can derive using calculus: For implicit F(x,y)=0, vertex lies along axis where gradient is parallel to axis direction. Better to give formula for axis-aligned parabola: A x^2 + D x + E y + F =0. Vertex: x_v = -D/(2A), y_v = - (A x_v^2 + D x_v + F)/E. If also rotated: need rotation. We can present general method: rotate by θ = ½ arctan(B/(A - C)). Then compute. Let's provide explicit formulas. Let A, B, C, D, E, F. Define θ = ½ arctan( B / (A - C) ) if A ≠ C, else θ = π/4. Rotate coordinates: x = x' cosθ - y' sinθ, y = x' sinθ + y' cosθ. Then coefficients transform. After rotation, B' =0, A' = A cos²θ + B sinθ cosθ + C sin²θ, C' = A sin²θ - B sinθ cosθ + C cos²θ. Parabola condition: A' C' =0. Assume C'=0, A'≠0. Then equation: A' x'^2 + D' x' + E' y' + F =0 where D' = D cosθ + E sinθ, E' = -D sinθ + E cosθ. Then vertex in primed coordinates: x'_v = -D'/(2 A'), y'_v = -(A' x'_v² + D' x'_v + F)/E'. Then rotate back. Might be too much. Probably user expects simple. Given typical high school: general form ax² + bx + c =0? Vertex at x = -b/(2a). I'll answer with both simple and general rotated. Provide formulas. Use nerdamer to verify? Might not need. Let's assume they want vertex of parabola y = ax² + bx + c. Then x_v = -b/(2a), y_v = c - b²/(4a). If they want general quadratic in x and y: Ax² + Bxy + Cy² + Dx + Ey + F =0. We can give formula for vertex when B=0. Probably sufficient. I'll answer with clear statement: For axis-parallel parabola. If they meant general conic, provide method. Given instruction to be critical and engage, ask clarifying. Provide answer. Let's output. Also could give formula using discriminant: For ax² + bx + c =0, vertex y = (4ac - b²)/(4a). Let's deliver. Final answer: For non-rotated: ... If rotated, etc. I'll give comprehensive. Let's also compute example with nerdamer to verify. { "name": "run_javascript", "arguments": { "code": "nerdamer('c - b^2/(4*a)').text('decimals')" } } [Tool Result] (no output) No output? Might need return. Let's just do simple. { "name": "run_javascript", "arguments": { "code": "const expr = nerdamer('c - b^2/(4*a)');\nconsole.log(expr.toString());" } } [Tool Result] (-1/4)*a^(-1)*b^2+c Good. Now answer. Probably provide final. For an axis-parallel parabola the “general cartesian form’’ is usually written * $y = a x^{2}+b x +c$, $a\ne 0$, or in implicit form * $a x^{2}+b x -\,e y+f =0$, $e\ne 0$, which is the same thing. Up to an inessential change of sign this is $a x^{2}+b x +c =0$ with $c\equiv c-e^{-1}f$ etc. Put $y=f(x)=a x^{2}+b x+c$. Then $$ f'(x)=2a x+b=0\iff x_{v}= -\frac{b}{2a} $$ and $$ y_{v}=f(x_{v})=a\frac{b^{2}}{4a^{2}}-b\frac{b}{2a}+c =c-\frac{b^{2}}{4a} =\frac{4ac-b^{2}}{4a}\; . $$ In implicit form $F(x,y)=a x^{2}+b x+e y+f=0$ with $e\ne0$ $$ x_{v}= -\frac{b}{2a},\qquad y_{v}= -\frac{a x_{v}^{2}+b x_{v}+f}{e} =-\frac{f}{e}+\frac{b^{2}}{4ae} $$ which is $y_{v}=c-b^{2}/4a$ for $c=-f/e$. In the usual $y=ax^{2}+bx+c$ notation this is the familiar * $h=x_{v}=-b/2a$, $k=y_{v}=c-b^{2}/4a$, so that $y-k=a(x-h)^{2}$ and the focus is $(h,k+p)$ with $p=1/4a$. *Remark on the fully rotated conic.* The completely general central form is $$ F(x,y)=Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0,\qquad B^{2}-4AC=0, $$ the $B^{2}-4AC=0$ being the parabola condition. Put $$ \cot 2\theta=\frac{A-C}{B},\qquad \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix} \begin{pmatrix}x'\\y'\end{pmatrix}, $$ so that $B'=0$. In the primed frame $$ A'x'^{2}+C'y'^{2}+D'x'+E'y'+F=0, $$ with $$ \begin{aligned} A'&=A\cos^{2}\theta+B\sin\theta\cos\theta+C\sin^{2}\theta,\\ C'&=A\sin^{2}\theta-B\sin\theta\cos\theta+C\cos^{2}\theta,\\ D'&=D\cos\theta+E\sin\theta,\\ E'&=-D\sin\theta+E\cos\theta, \end{aligned} $$ and $A'C'=0$ for a parabola. Up to relabelling we may take $C'=0\neq A'$. Then $$ A'x'^{2}+D'x'+E'y'+F=0\iff y'=-\frac{A'}{E'}x'^{2}-\frac{D'}{E'}x'-\frac{F}{E'}, $$ so that in the rotated system $$ x'_{v}=-\frac{D'}{2A'},\qquad y'_{v}= -\frac{A'x'_{v}{}^{2}+D'x'_{v}+F}{E'} . $$ The vertex in the original $x,y$ system is $$ \begin{pmatrix}x_{v}\\y_{v}\end{pmatrix} = \begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix} \begin{pmatrix}x'_{v}\\y'_{v}\end{pmatrix}, $$ i.e. $$ x_{v}=x'_{v}\cos\theta-y'_{v}\sin\theta,\qquad y_{v}=x'_{v}\sin\theta+y'_{v}\cos\theta . $$ For the usual non-rotated case $B=0$, $C=0$, $A\neq0$ this reduces to $x_{v}=-D/2A$, $y_{v}=-(A x_{v}^{2}+D x_{v}+F)/E$, which is the $-b/2a$, $c-b^{2}/4a$ formula above. The $A=0$, $C\neq0$ case is the same with the roles of $x$ and $y$ interchanged.