Find the Standard Form of the Parabola x=-(y-4)^2+3

$x = - {(y - 4)}^{2} + 3$

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- {(y - 4)}^{2} + 3 = x$

Subtract $3$ from both sides of the equation.

$- {(y - 4)}^{2} = x - 3$

Divide each term in $- {(y - 4)}^{2} = x - 3$ by $- 1$ and simplify.

Divide each term in $- {(y - 4)}^{2} = x - 3$ by $- 1$ .

$\frac{- {(y - 4)}^{2}}{- 1} = \frac{x}{- 1} + \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(y - 4)}^{2}}{1} = \frac{x}{- 1} + \frac{- 3}{- 1}$

Divide ${(y - 4)}^{2}$ by $1$ .

${(y - 4)}^{2} = \frac{x}{- 1} + \frac{- 3}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x}{- 1}$ .

${(y - 4)}^{2} = - 1 \cdot x + \frac{- 3}{- 1}$

Rewrite $- 1 \cdot x$ as $- x$ .

${(y - 4)}^{2} = - x + \frac{- 3}{- 1}$

Divide $- 3$ by $- 1$ .

${(y - 4)}^{2} = - x + 3$

Factor.

Factor $- 1$ out of $- x$ .

${(y - 4)}^{2} = - (x) + 3$

Rewrite $3$ as $- 1 (- 3)$ .

${(y - 4)}^{2} = - (x) - 1 (- 3)$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

${(y - 4)}^{2} = - (x - 3)$