Precalculus Examples

$x^{2} + y^{2} - x + 6 y + 9 = 0$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$x^{2} + {(0)}^{2} - x + 6 (0) + 9 = 0$

Step 1.2

Solve the equation.

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Step 1.2.1

Simplify $x^{2} + {(0)}^{2} - x + 6 (0) + 9$ .

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Step 1.2.1.1

Simplify each term.

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Step 1.2.1.1.1

Raising $0$ to any positive power yields $0$ .

$x^{2} + 0 - x + 6 (0) + 9 = 0$

Step 1.2.1.1.2

Multiply $6$ by $0$ .

$x^{2} + 0 - x + 0 + 9 = 0$

Step 1.2.1.2

Combine the opposite terms in $x^{2} + 0 - x + 0 + 9$ .

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Step 1.2.1.2.1

Add $x^{2}$ and $0$ .

$x^{2} - x + 0 + 9 = 0$

Step 1.2.1.2.2

Add $x^{2} - x$ and $0$ .

$x^{2} - x + 9 = 0$

Step 1.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 1.2.3

Substitute the values $a = 1$ , $b = - 1$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Step 1.2.4

Simplify.

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Step 1.2.4.1

Simplify the numerator.

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Step 1.2.4.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Step 1.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 1.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 9}}{2 \cdot 1}$

Step 1.2.4.1.2.2

Multiply $- 4$ by $9$ .

$x = \frac{1 \pm \sqrt{1 - 36}}{2 \cdot 1}$

Step 1.2.4.1.3

Subtract $36$ from $1$ .

$x = \frac{1 \pm \sqrt{- 35}}{2 \cdot 1}$

Step 1.2.4.1.4

Rewrite $- 35$ as $- 1 (35)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 35}}{2 \cdot 1}$

Step 1.2.4.1.5

Rewrite $\sqrt{- 1 (35)}$ as $\sqrt{- 1} \cdot \sqrt{35}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{35}}{2 \cdot 1}$

Step 1.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{35}}{2 \cdot 1}$

Step 1.2.4.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{35}}{2}$

Step 1.2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.2.5.1

Simplify the numerator.

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Step 1.2.5.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Step 1.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 1.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 9}}{2 \cdot 1}$

Step 1.2.5.1.2.2

Multiply $- 4$ by $9$ .

$x = \frac{1 \pm \sqrt{1 - 36}}{2 \cdot 1}$

Step 1.2.5.1.3

Subtract $36$ from $1$ .

$x = \frac{1 \pm \sqrt{- 35}}{2 \cdot 1}$

Step 1.2.5.1.4

Rewrite $- 35$ as $- 1 (35)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 35}}{2 \cdot 1}$

Step 1.2.5.1.5

Rewrite $\sqrt{- 1 (35)}$ as $\sqrt{- 1} \cdot \sqrt{35}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{35}}{2 \cdot 1}$

Step 1.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{35}}{2 \cdot 1}$

Step 1.2.5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{35}}{2}$

Step 1.2.5.3

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{35}}{2}$

Step 1.2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.2.6.1

Simplify the numerator.

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Step 1.2.6.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Step 1.2.6.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 1.2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 9}}{2 \cdot 1}$

Step 1.2.6.1.2.2

Multiply $- 4$ by $9$ .

$x = \frac{1 \pm \sqrt{1 - 36}}{2 \cdot 1}$

Step 1.2.6.1.3

Subtract $36$ from $1$ .

$x = \frac{1 \pm \sqrt{- 35}}{2 \cdot 1}$

Step 1.2.6.1.4

Rewrite $- 35$ as $- 1 (35)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 35}}{2 \cdot 1}$

Step 1.2.6.1.5

Rewrite $\sqrt{- 1 (35)}$ as $\sqrt{- 1} \cdot \sqrt{35}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{35}}{2 \cdot 1}$

Step 1.2.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{35}}{2 \cdot 1}$

Step 1.2.6.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{35}}{2}$

Step 1.2.6.3

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{35}}{2}$

Step 1.2.7

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{35}}{2}, \frac{1 - i \sqrt{35}}{2}$

Step 1.3

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

x-intercept(s): $None$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

${(0)}^{2} + y^{2} - (0) + 6 y + 9 = 0$

Step 2.2

Solve the equation.

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Step 2.2.1

Simplify ${(0)}^{2} + y^{2} - (0) + 6 y + 9$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Raising $0$ to any positive power yields $0$ .

$0 + y^{2} - (0) + 6 y + 9 = 0$

Step 2.2.1.1.2

Multiply $- 1$ by $0$ .

$0 + y^{2} + 0 + 6 y + 9 = 0$

Step 2.2.1.2

Combine the opposite terms in $0 + y^{2} + 0 + 6 y + 9$ .

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Step 2.2.1.2.1

Add $0$ and $y^{2}$ .

$y^{2} + 0 + 6 y + 9 = 0$

Step 2.2.1.2.2

Add $y^{2}$ and $0$ .

$y^{2} + 6 y + 9 = 0$

Step 2.2.2

Factor using the perfect square rule.

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Step 2.2.2.1

Rewrite $9$ as $3^{2}$ .

$y^{2} + 6 y + 3^{2} = 0$

Step 2.2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 y = 2 \cdot y \cdot 3$

Step 2.2.2.3

Rewrite the polynomial.

$y^{2} + 2 \cdot y \cdot 3 + 3^{2} = 0$

Step 2.2.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = y$ and $b = 3$ .

${(y + 3)}^{2} = 0$

Step 2.2.3

Set the $y + 3$ equal to $0$ .

$y + 3 = 0$

Step 2.2.4

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

Step 3

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, - 3)$

Step 4