Find the Standard Form of the Parabola y^2+14y+16x+33=0

$y^{2} + 14 y + 16 x + 33 = 0$

Step 1

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

Subtract $16 x$ from both sides of the equation.

$y^{2} + 14 y + 33 = - 16 x$

Subtract $33$ from both sides of the equation.

$y^{2} + 14 y = - 16 x - 33$

Step 2

Complete the square.

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

$a = 1$

$b = 14$

$c = 0 = - 16 x - 33$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = - 16 x - 33$

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

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

$d = \frac{14}{2 \cdot 1}$

Cancel the common factor of $14$ and $2$ .

Factor $2$ out of $14$ .

$d = \frac{2 \cdot 7}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 7}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 7}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{7}{1}$

Divide $7$ by $1$ .

$d = 7$

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

$e = 0 - \frac{14^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $14$ to the power of $2$ .

$e = 0 - \frac{196}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{196}{4}$

Divide $196$ by $4$ .

$e = 0 - 1 \cdot 49$

Multiply $- 1$ by $49$ .

$e = 0 - 49$

Subtract $49$ from $0$ .

$e = - 49$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + 7)}^{2} - 49$ .

${(y + 7)}^{2} - 49 = - 16 x - 33$

Step 3

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

Add $49$ to both sides of the equation.

${(y + 7)}^{2} = - 16 x - 33 + 49$

Add $- 33$ and $49$ .

${(y + 7)}^{2} = - 16 x + 16$

Step 4

Factor.

Factor $16$ out of $- 16 x + 16$ .

Factor $16$ out of $- 16 x$ .

${(y + 7)}^{2} = 16 (- x) + 16$

Factor $16$ out of $16$ .

${(y + 7)}^{2} = 16 (- x) + 16 (1)$

Factor $16$ out of $16 (- x) + 16 (1)$ .

${(y + 7)}^{2} = 16 (- x + 1)$

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

${(y + 7)}^{2} = 16 (- (x) + 1)$

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

${(y + 7)}^{2} = 16 (- (x) - 1 (- 1))$

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

${(y + 7)}^{2} = 16 (- (x - 1))$

Multiply $- 1$ by $16$ .

${(y + 7)}^{2} = - 16 (x - 1)$

Find the Standard Form of the Parabola y^2+14y+16x+33=0

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