Find the Standard Form of the Parabola h(x)=-5(x-1)^2+3

$h (x) = - 5 {(x - 1)}^{2} + 3$

Step 1

Write $h (x) = - 5 {(x - 1)}^{2} + 3$ as an equation.

$y = - 5 {(x - 1)}^{2} + 3$

Step 2

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

$- 5 {(x - 1)}^{2} + 3 = y$

Step 3

Subtract $3$ from both sides of the equation.

$- 5 {(x - 1)}^{2} = y - 3$

Step 4

Divide each term in $- 5 {(x - 1)}^{2} = y - 3$ by $- 5$ and simplify.

Divide each term in $- 5 {(x - 1)}^{2} = y - 3$ by $- 5$ .

$\frac{- 5 {(x - 1)}^{2}}{- 5} = \frac{y}{- 5} + \frac{- 3}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 {(x - 1)}^{2}}{- 5} = \frac{y}{- 5} + \frac{- 3}{- 5}$

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

${(x - 1)}^{2} = \frac{y}{- 5} + \frac{- 3}{- 5}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

${(x - 1)}^{2} = - \frac{y}{5} + \frac{- 3}{- 5}$

Dividing two negative values results in a positive value.

${(x - 1)}^{2} = - \frac{y}{5} + \frac{3}{5}$

Step 5

Factor.

Combine the numerators over the common denominator.

${(x - 1)}^{2} = \frac{- y + 3}{5}$

Factor $- 1$ out of $- y$ .

${(x - 1)}^{2} = \frac{- (y) + 3}{5}$

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

${(x - 1)}^{2} = \frac{- (y) - 1 (- 3)}{5}$

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

${(x - 1)}^{2} = \frac{- (y - 3)}{5}$

Move the negative in front of the fraction.

${(x - 1)}^{2} = - \frac{y - 3}{5}$

Reorder terms.

${(x - 1)}^{2} = - (\frac{1}{5} (y - 3))$

Remove parentheses.

${(x - 1)}^{2} = - \frac{1}{5} (y - 3)$