Solve y=m(x-3)^2

$$y = m ( x - 3 ) ^ { 2 }$$

Solve for m

$$\left\{\begin{matrix}m=\frac{y}{\left(x-3\right)^{2}}\text{, }&x\neq 3\\m\in \mathrm{R}\text{, }&y=0\text{ and }x=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$y=m\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply m by x^{2}-6x+9.

$$y=mx^{2}-6mx+9m$$

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

$$mx^{2}-6mx+9m=y$$

Combine all terms containing m.

$$\left(x^{2}-6x+9\right)m=y$$

Divide both sides by x^{2}-6x+9.

$$\frac{\left(x^{2}-6x+9\right)m}{x^{2}-6x+9}=\frac{y}{x^{2}-6x+9}$$

Dividing by x^{2}-6x+9 undoes the multiplication by x^{2}-6x+9.

$$m=\frac{y}{x^{2}-6x+9}$$

Divide y by x^{2}-6x+9.

$$m=\frac{y}{\left(x-3\right)^{2}}$$

Solve for x

$$\left\{\begin{matrix}x=-\sqrt{\frac{y}{m}}+3\text{; }x=\sqrt{\frac{y}{m}}+3\text{, }&y\leq 0\text{ and }m<0\\x=-\sqrt{\frac{y}{m}}+3\text{; }x=\sqrt{\frac{y}{m}}+3\text{, }&y\geq 0\text{ and }m>0\\x\in \mathrm{R}\text{, }&y=0\text{ and }m=0\end{matrix}\right.$$