Precalculus Examples

$16 x^{2} + 80 x + 16 y^{2} - 112 y + 247 = 0$

Step 1

Subtract $247$ from both sides of the equation.

$16 x^{2} + 80 x + 16 y^{2} - 112 y = - 247$

Step 2

Divide both sides of the equation by $16$ .

$x^{2} + 5 x + y^{2} - 7 y = - \frac{247}{16}$

Step 3

Complete the square for $x^{2} + 5 x$ .

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Step 3.1

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

$a = 1$

$b = 5$

$c = 0$

Step 3.2

Consider the vertex form of a parabola.

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

Step 3.3

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

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Step 3.3.1

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

$d = \frac{5}{2 \cdot 1}$

Step 3.3.2

Multiply $2$ by $1$ .

$d = \frac{5}{2}$

Step 3.4

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

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Step 3.4.1

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

$e = 0 - \frac{5^{2}}{4 \cdot 1}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Raise $5$ to the power of $2$ .

$e = 0 - \frac{25}{4 \cdot 1}$

Step 3.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{25}{4}$

Step 3.4.2.2

Subtract $\frac{25}{4}$ from $0$ .

$e = - \frac{25}{4}$

Step 3.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{5}{2})}^{2} - \frac{25}{4}$ .

${(x + \frac{5}{2})}^{2} - \frac{25}{4}$

Step 4

Substitute ${(x + \frac{5}{2})}^{2} - \frac{25}{4}$ for $x^{2} + 5 x$ in the equation $x^{2} + 5 x + y^{2} - 7 y = - \frac{247}{16}$ .

${(x + \frac{5}{2})}^{2} - \frac{25}{4} + y^{2} - 7 y = - \frac{247}{16}$

Step 5

Move $- \frac{25}{4}$ to the right side of the equation by adding $\frac{25}{4}$ to both sides.

${(x + \frac{5}{2})}^{2} + y^{2} - 7 y = - \frac{247}{16} + \frac{25}{4}$

Step 6

Complete the square for $y^{2} - 7 y$ .

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Step 6.1

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

$a = 1$

$b = - 7$

$c = 0$

Step 6.2

Consider the vertex form of a parabola.

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

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Multiply $2$ by $1$ .

$d = \frac{- 7}{2}$

Step 6.3.2.2

Move the negative in front of the fraction.

$d = - \frac{7}{2}$

Step 6.4

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

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Step 6.4.1

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

$e = 0 - \frac{{(- 7)}^{2}}{4 \cdot 1}$

Step 6.4.2

Simplify the right side.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

Raise $- 7$ to the power of $2$ .

$e = 0 - \frac{49}{4 \cdot 1}$

Step 6.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{49}{4}$

Step 6.4.2.2

Subtract $\frac{49}{4}$ from $0$ .

$e = - \frac{49}{4}$

Step 6.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - \frac{7}{2})}^{2} - \frac{49}{4}$ .

${(y - \frac{7}{2})}^{2} - \frac{49}{4}$

Step 7

Substitute ${(y - \frac{7}{2})}^{2} - \frac{49}{4}$ for $y^{2} - 7 y$ in the equation $x^{2} + 5 x + y^{2} - 7 y = - \frac{247}{16}$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} - \frac{49}{4} = - \frac{247}{16} + \frac{25}{4}$

Step 8

Move $- \frac{49}{4}$ to the right side of the equation by adding $\frac{49}{4}$ to both sides.

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{25}{4} + \frac{49}{4}$

Step 9

Simplify $- \frac{247}{16} + \frac{25}{4} + \frac{49}{4}$ .

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Step 9.1

Combine fractions.

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Step 9.1.1

Combine the numerators over the common denominator.

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{- 247}{16} + \frac{25 + 49}{4}$

Step 9.1.2

Add $25$ and $49$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{- 247}{16} + \frac{74}{4}$

Step 9.2

Simplify each term.

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Step 9.2.1

Move the negative in front of the fraction.

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{74}{4}$

Step 9.2.2

Cancel the common factor of $74$ and $4$ .

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Step 9.2.2.1

Factor $2$ out of $74$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{2 (37)}{4}$

Step 9.2.2.2

Cancel the common factors.

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Step 9.2.2.2.1

Factor $2$ out of $4$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{2 \cdot 37}{2 \cdot 2}$

Step 9.2.2.2.2

Cancel the common factor.

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{2 \cdot 37}{2 \cdot 2}$

Step 9.2.2.2.3

Rewrite the expression.

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{37}{2}$

Step 9.3

To write $\frac{37}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{37}{2} \cdot \frac{8}{8}$

Step 9.4

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Step 9.4.1

Multiply $\frac{37}{2}$ by $\frac{8}{8}$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{37 \cdot 8}{2 \cdot 8}$

Step 9.4.2

Multiply $2$ by $8$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = - \frac{247}{16} + \frac{37 \cdot 8}{16}$

Step 9.5

Combine the numerators over the common denominator.

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{- 247 + 37 \cdot 8}{16}$

Step 9.6

Simplify the numerator.

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Step 9.6.1

Multiply $37$ by $8$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{- 247 + 296}{16}$

Step 9.6.2

Add $- 247$ and $296$ .

${(x + \frac{5}{2})}^{2} + {(y - \frac{7}{2})}^{2} = \frac{49}{16}$

Step 10

This is the form of a circle. Use this form to determine the center and radius of the circle.

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Step 11

Match the values in this circle to those of the standard form. The variable $r$ represents the radius of the circle, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from origin.

$r = \frac{7}{4}$

$h = - \frac{5}{2}$

$k = \frac{7}{2}$

Step 12

The center of the circle is found at $(h, k)$ .

Center: $(- \frac{5}{2}, \frac{7}{2})$

Step 13

These values represent the important values for graphing and analyzing a circle.

Center: $(- \frac{5}{2}, \frac{7}{2})$

Radius: $\frac{7}{4}$

Step 14