Find the x and y Intercepts x^2-8x+y^2+6y=-9

$x^{2} - 8 x + y^{2} + 6 y = - 9$

Find the x-intercepts.

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

$x^{2} - 8 x + {(0)}^{2} + 6 (0) = - 9$

Solve the equation.

Simplify $x^{2} - 8 x + {(0)}^{2} + 6 (0)$ .

Simplify each term.

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

$x^{2} - 8 x + 0 + 6 (0) = - 9$

Multiply $6$ by $0$ .

$x^{2} - 8 x + 0 + 0 = - 9$

Combine the opposite terms in $x^{2} - 8 x + 0 + 0$ .

Add $x^{2} - 8 x$ and $0$ .

$x^{2} - 8 x + 0 = - 9$

Add $x^{2} - 8 x$ and $0$ .

$x^{2} - 8 x = - 9$

Add $9$ to both sides of the equation.

$x^{2} - 8 x + 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 8$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{8 \pm \sqrt{64 - 36}}{2 \cdot 1}$

Subtract $36$ from $64$ .

$x = \frac{8 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{8 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{8 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{8 \pm 2 \sqrt{7}}{2}$ .

$x = 4 \pm \sqrt{7}$

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

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{8 \pm \sqrt{64 - 36}}{2 \cdot 1}$

Subtract $36$ from $64$ .

$x = \frac{8 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{8 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{8 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{8 \pm 2 \sqrt{7}}{2}$ .

$x = 4 \pm \sqrt{7}$

Change the $\pm$ to $+$ .

$x = 4 + \sqrt{7}$

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

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{8 \pm \sqrt{64 - 36}}{2 \cdot 1}$

Subtract $36$ from $64$ .

$x = \frac{8 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{8 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{8 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{8 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{8 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{8 \pm 2 \sqrt{7}}{2}$ .

$x = 4 \pm \sqrt{7}$

Change the $\pm$ to $-$ .

$x = 4 - \sqrt{7}$

The final answer is the combination of both solutions.

$x = 4 + \sqrt{7}, 4 - \sqrt{7}$

x-intercept(s) in point form.

x-intercept(s): $(4 + \sqrt{7}, 0), (4 - \sqrt{7}, 0)$

Find the y-intercepts.

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

${(0)}^{2} - 8 \cdot 0 + y^{2} + 6 y = - 9$

Solve the equation.

Simplify ${(0)}^{2} - 8 \cdot 0 + y^{2} + 6 y$ .

Simplify each term.

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

$0 - 8 \cdot 0 + y^{2} + 6 y = - 9$

Multiply $- 8$ by $0$ .

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

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

Add $0$ and $0$ .

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

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

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

Add $9$ to both sides of the equation.

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

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 $0$ .

$y + 3 = 0$

Subtract $3$ from both sides of the equation.

$y = - 3$

y-intercept(s) in point form.

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

List the intersections.

x-intercept(s): $(4 + \sqrt{7}, 0), (4 - \sqrt{7}, 0)$

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

Find the x and y Intercepts x^2-8x+y^2+6y=-9

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