Find the Standard Form of the Parabola y^2-2y+12x-35=0

$y^{2} - 2 y + 12 x - 35 = 0$

Step 1

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

Subtract $12 x$ from both sides of the equation.

$y^{2} - 2 y - 35 = - 12 x$

Add $35$ to both sides of the equation.

$y^{2} - 2 y = - 12 x + 35$

Step 2

Complete the square.

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

$a = 1$

$b = - 2$

$c = 0 = - 12 x + 35$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - 12 x + 35$

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 $- 2$ and $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot - 1}{2 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{- 1}{1}$

Divide $- 1$ by $1$ .

$d = - 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 2)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = 0 - \frac{4}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = 0 - \frac{4}{4}$

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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

${(y - 1)}^{2} - 1 = - 12 x + 35$

Step 3

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

Add $1$ to both sides of the equation.

${(y - 1)}^{2} = - 12 x + 35 + 1$

Add $35$ and $1$ .

${(y - 1)}^{2} = - 12 x + 36$

Step 4

Factor.

Factor $12$ out of $- 12 x + 36$ .

Factor $12$ out of $- 12 x$ .

${(y - 1)}^{2} = 12 (- x) + 36$

Factor $12$ out of $36$ .

${(y - 1)}^{2} = 12 (- x) + 12 (3)$

Factor $12$ out of $12 (- x) + 12 (3)$ .

${(y - 1)}^{2} = 12 (- x + 3)$

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

${(y - 1)}^{2} = 12 (- (x) + 3)$

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

${(y - 1)}^{2} = 12 (- (x) - 1 (- 3))$

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

${(y - 1)}^{2} = 12 (- (x - 3))$

Multiply $- 1$ by $12$ .

${(y - 1)}^{2} = - 12 (x - 3)$

Find the Standard Form of the Parabola y^2-2y+12x-35=0

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