Find the x and y Intercepts x^2+y^2-4x-2y-4=0

$x^{2} + y^{2} - 4 x - 2 y - 4 = 0$

Step 1

Find the x-intercepts.

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

$x^{2} + {(0)}^{2} - 4 x - 2 \cdot 0 - 4 = 0$

Solve the equation.

Simplify $x^{2} + {(0)}^{2} - 4 x - 2 \cdot 0 - 4$ .

Simplify each term.

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

$x^{2} + 0 - 4 x - 2 \cdot 0 - 4 = 0$

Multiply $- 2$ by $0$ .

$x^{2} + 0 - 4 x + 0 - 4 = 0$

Combine the opposite terms in $x^{2} + 0 - 4 x + 0 - 4$ .

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

$x^{2} - 4 x + 0 - 4 = 0$

Add $x^{2} - 4 x$ and $0$ .

$x^{2} - 4 x - 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 + 16}}{2 \cdot 1}$

Add $16$ and $16$ .

$x = \frac{4 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{4 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

$x = \frac{4 \pm 4 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{2}}{2}$ .

$x = 2 \pm 2 \sqrt{2}$

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

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 + 16}}{2 \cdot 1}$

Add $16$ and $16$ .

$x = \frac{4 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{4 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

$x = \frac{4 \pm 4 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{2}}{2}$ .

$x = 2 \pm 2 \sqrt{2}$

Change the $\pm$ to $+$ .

$x = 2 + 2 \sqrt{2}$

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

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 4}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 4}}{2 \cdot 1}$

Multiply $- 4$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 + 16}}{2 \cdot 1}$

Add $16$ and $16$ .

$x = \frac{4 \pm \sqrt{32}}{2 \cdot 1}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$x = \frac{4 \pm \sqrt{16 (2)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

$x = \frac{4 \pm 4 \sqrt{2}}{2}$

Simplify $\frac{4 \pm 4 \sqrt{2}}{2}$ .

$x = 2 \pm 2 \sqrt{2}$

Change the $\pm$ to $-$ .

$x = 2 - 2 \sqrt{2}$

The final answer is the combination of both solutions.

$x = 2 + 2 \sqrt{2}, 2 - 2 \sqrt{2}$

x-intercept(s) in point form.

x-intercept(s): $(2 + 2 \sqrt{2}, 0), (2 - 2 \sqrt{2}, 0)$

Step 2

Find the y-intercepts.

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

${(0)}^{2} + y^{2} - 4 \cdot 0 - 2 y - 4 = 0$

Solve the equation.

Simplify ${(0)}^{2} + y^{2} - 4 \cdot 0 - 2 y - 4$ .

Simplify each term.

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

$0 + y^{2} - 4 \cdot 0 - 2 y - 4 = 0$

Multiply $- 4$ by $0$ .

$0 + y^{2} + 0 - 2 y - 4 = 0$

Combine the opposite terms in $0 + y^{2} + 0 - 2 y - 4$ .

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

$y^{2} + 0 - 2 y - 4 = 0$

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

$y^{2} - 2 y - 4 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 4$ into the quadratic formula and solve for $y$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$y = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$y = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$y = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$y = 1 \pm \sqrt{5}$

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$y = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$y = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$y = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$y = 1 \pm \sqrt{5}$

Change the $\pm$ to $+$ .

$y = 1 + \sqrt{5}$

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 + 16}}{2 \cdot 1}$

Add $4$ and $16$ .

$y = \frac{2 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$y = \frac{2 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$y = \frac{2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$y = 1 \pm \sqrt{5}$

Change the $\pm$ to $-$ .

$y = 1 - \sqrt{5}$

The final answer is the combination of both solutions.

$y = 1 + \sqrt{5}, 1 - \sqrt{5}$

y-intercept(s) in point form.

y-intercept(s): $(0, 1 + \sqrt{5}), (0, 1 - \sqrt{5})$

Step 3

List the intersections.

x-intercept(s): $(2 + 2 \sqrt{2}, 0), (2 - 2 \sqrt{2}, 0)$

y-intercept(s): $(0, 1 + \sqrt{5}), (0, 1 - \sqrt{5})$

Step 4

Find the x and y Intercepts x^2+y^2-4x-2y-4=0

Related Questions

Support LCMGCF.com by disabling your adblocker.