Solve y=-a(x-b)^2+c

$$y = - a ( x - b ) ^ { 2 } + c$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{c-y}{\left(x-b\right)^{2}}\text{, }&x\neq b\\a\in \mathrm{C}\text{, }&y=c\text{ and }x=b\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(x-b\right)^{2}.

$$y=\left(-a\right)\left(x^{2}-2xb+b^{2}\right)+c$$

Use the distributive property to multiply -a by x^{2}-2xb+b^{2}.

$$y=\left(-a\right)x^{2}-2\left(-a\right)xb+\left(-a\right)b^{2}+c$$

Multiply -2 and -1 to get 2.

$$y=\left(-a\right)x^{2}+2axb+\left(-a\right)b^{2}+c$$

Swap sides so that all variable terms are on the left hand side.

$$\left(-a\right)x^{2}+2axb+\left(-a\right)b^{2}+c=y$$

Subtract c from both sides.

$$\left(-a\right)x^{2}+2axb+\left(-a\right)b^{2}=y-c$$

Reorder the terms.

$$-ax^{2}+2abx-ab^{2}=y-c$$

Combine all terms containing a.

$$\left(-x^{2}+2bx-b^{2}\right)a=y-c$$

Divide both sides by -x^{2}+2bx-b^{2}.

$$\frac{\left(-x^{2}+2bx-b^{2}\right)a}{-x^{2}+2bx-b^{2}}=\frac{y-c}{-x^{2}+2bx-b^{2}}$$

Dividing by -x^{2}+2bx-b^{2} undoes the multiplication by -x^{2}+2bx-b^{2}.

$$a=\frac{y-c}{-x^{2}+2bx-b^{2}}$$

Divide y-c by -x^{2}+2bx-b^{2}.

$$a=-\frac{y-c}{\left(x-b\right)^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{c-y}{\left(x-b\right)^{2}}\text{, }&x\neq b\\a\in \mathrm{R}\text{, }&y=c\text{ and }x=b\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(x-b\right)^{2}.

$$y=\left(-a\right)\left(x^{2}-2xb+b^{2}\right)+c$$

Use the distributive property to multiply -a by x^{2}-2xb+b^{2}.

$$y=\left(-a\right)x^{2}-2\left(-a\right)xb+\left(-a\right)b^{2}+c$$

Multiply -2 and -1 to get 2.

$$y=\left(-a\right)x^{2}+2axb+\left(-a\right)b^{2}+c$$

Swap sides so that all variable terms are on the left hand side.

$$\left(-a\right)x^{2}+2axb+\left(-a\right)b^{2}+c=y$$

Subtract c from both sides.

$$\left(-a\right)x^{2}+2axb+\left(-a\right)b^{2}=y-c$$

Reorder the terms.

$$-ax^{2}+2abx-ab^{2}=y-c$$

Combine all terms containing a.

$$\left(-x^{2}+2bx-b^{2}\right)a=y-c$$

Divide both sides by -x^{2}+2bx-b^{2}.

$$\frac{\left(-x^{2}+2bx-b^{2}\right)a}{-x^{2}+2bx-b^{2}}=\frac{y-c}{-x^{2}+2bx-b^{2}}$$

Dividing by -x^{2}+2bx-b^{2} undoes the multiplication by -x^{2}+2bx-b^{2}.

$$a=\frac{y-c}{-x^{2}+2bx-b^{2}}$$

Divide y-c by -x^{2}+2bx-b^{2}.

$$a=-\frac{y-c}{\left(x-b\right)^{2}}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=x-ia^{-\frac{1}{2}}\sqrt{y-c}\text{; }b=x+ia^{-\frac{1}{2}}\sqrt{y-c}\text{, }&a\neq 0\\b\in \mathrm{C}\text{, }&y=c\text{ and }a=0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b=\frac{\sqrt{a}x+\sqrt{c-y}}{\sqrt{a}}\text{; }b=\frac{\sqrt{a}x-\sqrt{c-y}}{\sqrt{a}}\text{, }&a>0\text{ and }y\leq c\\b=x+\sqrt{-\frac{y-c}{a}}\text{; }b=x-\sqrt{-\frac{y-c}{a}}\text{, }&y\geq c\text{ and }a<0\\b=x\text{, }&y=c\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&y=c\text{ and }a=0\end{matrix}\right.$$