Precalculus Examples

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

Step 1

Subtract $3 y$ from both sides of the equation.

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

Step 2

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

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

$a = 1$

$b = - 3$

$c = 0$

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 $d = \frac{b}{2 a}$ .

$d = \frac{- 3}{2 \cdot 1}$

Simplify the right side.

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Multiply $2$ by $1$ .

$d = \frac{- 3}{2}$

Move the negative in front of the fraction.

$d = - \frac{3}{2}$

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

$e = 0 - \frac{{(- 3)}^{2}}{4 \cdot 1}$

Simplify the right side.

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Simplify each term.

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Raise $- 3$ to the power of $2$ .

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

Multiply $4$ by $1$ .

$e = 0 - \frac{9}{4}$

Subtract $\frac{9}{4}$ from $0$ .

$e = - \frac{9}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - \frac{3}{2})}^{2} - \frac{9}{4}$ .

${(y - \frac{3}{2})}^{2} - \frac{9}{4}$

Step 3

Substitute ${(y - \frac{3}{2})}^{2} - \frac{9}{4}$ for $y^{2} - 3 y$ in the equation $x^{2} + y^{2} - 3 y = 0$ .

$x^{2} + {(y - \frac{3}{2})}^{2} - \frac{9}{4} = 0$

Step 4

Move $- \frac{9}{4}$ to the right side of the equation by adding $\frac{9}{4}$ to both sides.

$x^{2} + {(y - \frac{3}{2})}^{2} = 0 + \frac{9}{4}$

Step 5

Add $0$ and $\frac{9}{4}$ .

$x^{2} + {(y - \frac{3}{2})}^{2} = \frac{9}{4}$