Find the Vertex Form 0=y^2-x+2y-3

$0 = y^{2} - x + 2 y - 3$

Step 1

Isolate $x$ to the left side of the equation.

Rewrite the equation as $y^{2} - x + 2 y - 3 = 0$ .

$y^{2} - x + 2 y - 3 = 0$

Move all terms not containing $x$ to the right side of the equation.

Subtract $y^{2}$ from both sides of the equation.

$- x + 2 y - 3 = - y^{2}$

Subtract $2 y$ from both sides of the equation.

$- x - 3 = - y^{2} - 2 y$

Add $3$ to both sides of the equation.

$- x = - y^{2} - 2 y + 3$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

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

Divide $y^{2}$ by $1$ .

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

Move the negative one from the denominator of $\frac{- 2 y}{- 1}$ .

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

Rewrite $- 1 \cdot (- 2 y)$ as $- (- 2 y)$ .

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

Multiply $- 2$ by $- 1$ .

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

Divide $3$ by $- 1$ .

$x = y^{2} + 2 y - 3$

Step 2

Complete the square for $y^{2} + 2 y - 3$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 2$

$c = - 3$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{2}{2 \cdot 1}$

Cancel the common factor of $2$ .

Cancel the common factor.

$d = \frac{2}{2 \cdot 1}$

Rewrite the expression.

$d = 1$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = - 3 - \frac{2^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $2$ to the power of $2$ .

$e = - 3 - \frac{4}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = - 3 - \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = - 3 - \frac{4}{4}$

Rewrite the expression.

$e = - 3 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = - 3 - 1$

Subtract $1$ from $- 3$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 1)}^{2} - 4$ .

${(y + 1)}^{2} - 4$

Step 3

Set $x$ equal to the new right side.

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

Find the Vertex Form 0=y^2-x+2y-3

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