Find the x and y Intercepts x^2+y^2-3x-6y+9=0

$x^{2} + y^{2} - 3 x - 6 y + 9 = 0$

Step 1

Find the x-intercepts.

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

$x^{2} + {(0)}^{2} - 3 x - 6 \cdot 0 + 9 = 0$

Solve the equation.

Simplify $x^{2} + {(0)}^{2} - 3 x - 6 \cdot 0 + 9$ .

Simplify each term.

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

$x^{2} + 0 - 3 x - 6 \cdot 0 + 9 = 0$

Multiply $- 6$ by $0$ .

$x^{2} + 0 - 3 x + 0 + 9 = 0$

Combine the opposite terms in $x^{2} + 0 - 3 x + 0 + 9$ .

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

$x^{2} - 3 x + 0 + 9 = 0$

Add $x^{2} - 3 x$ and $0$ .

$x^{2} - 3 x + 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 3$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$x = \frac{3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$x = \frac{3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$x = \frac{3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

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

$x = \frac{3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Move $3$ to the left of $i$ .

$x = \frac{3 \pm 3 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm 3 i \sqrt{3}}{2}$

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

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$x = \frac{3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$x = \frac{3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$x = \frac{3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

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

$x = \frac{3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Move $3$ to the left of $i$ .

$x = \frac{3 \pm 3 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm 3 i \sqrt{3}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{3 + 3 i \sqrt{3}}{2}$

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

Simplify the numerator.

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

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$x = \frac{3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$x = \frac{3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$x = \frac{3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

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

$x = \frac{3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Move $3$ to the left of $i$ .

$x = \frac{3 \pm 3 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{3 \pm 3 i \sqrt{3}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{3 - 3 i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = \frac{3 + 3 i \sqrt{3}}{2}, \frac{3 - 3 i \sqrt{3}}{2}$

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.

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

${(0)}^{2} + y^{2} - 3 \cdot 0 - 6 y + 9 = 0$

Solve the equation.

Simplify ${(0)}^{2} + y^{2} - 3 \cdot 0 - 6 y + 9$ .

Simplify each term.

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

$0 + y^{2} - 3 \cdot 0 - 6 y + 9 = 0$

Multiply $- 3$ by $0$ .

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

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

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

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

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

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

Factor using the perfect square rule.

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

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

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

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

Rewrite the polynomial.

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

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

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

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

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

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

Find the x and y Intercepts x^2+y^2-3x-6y+9=0

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