Precalculus Examples

$4 x^{2} + 4 y^{2} + 36 y + 5 = 0$

Subtract $5$ from both sides of the equation.

$4 x^{2} + 4 y^{2} + 36 y = - 5$

Divide both sides of the equation by $4$ .

$x^{2} + y^{2} + 9 y = - \frac{5}{4}$

Complete the square for $y^{2} + 9 y$ .

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

$a = 1$

$b = 9$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{9}{2 \cdot 1}$

Multiply $2$ by $1$ .

$d = \frac{9}{2}$

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

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

$e = 0 - \frac{9^{2}}{4 \cdot 1}$

Simplify the right side.

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Simplify each term.

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Raise $9$ to the power of $2$ .

$e = 0 - \frac{81}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{81}{4}$

Subtract $\frac{81}{4}$ from $0$ .

$e = - \frac{81}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y + \frac{9}{2})}^{2} - \frac{81}{4}$ .

${(y + \frac{9}{2})}^{2} - \frac{81}{4}$

Substitute ${(y + \frac{9}{2})}^{2} - \frac{81}{4}$ for $y^{2} + 9 y$ in the equation $x^{2} + y^{2} + 9 y = - \frac{5}{4}$ .

$x^{2} + {(y + \frac{9}{2})}^{2} - \frac{81}{4} = - \frac{5}{4}$

Move $- \frac{81}{4}$ to the right side of the equation by adding $\frac{81}{4}$ to both sides.

$x^{2} + {(y + \frac{9}{2})}^{2} = - \frac{5}{4} + \frac{81}{4}$

Simplify $- \frac{5}{4} + \frac{81}{4}$ .

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Combine the numerators over the common denominator.

$x^{2} + {(y + \frac{9}{2})}^{2} = \frac{- 5 + 81}{4}$

Simplify the expression.

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Add $- 5$ and $81$ .

$x^{2} + {(y + \frac{9}{2})}^{2} = \frac{76}{4}$

Divide $76$ by $4$ .

$x^{2} + {(y + \frac{9}{2})}^{2} = 19$