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

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

Find the x-intercepts.

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

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

Solve the equation.

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

Simplify each term.

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

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

Multiply $4$ by $0$ .

$x^{2} + 0 = 4$

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

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

x-intercept(s) in point form.

x-intercept(s): $(2, 0), (- 2, 0)$

Find the y-intercepts.

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

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

Solve the equation.

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

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

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

Add $0$ and $4 y^{2}$ .

$4 y^{2} = 4$

Divide each term in $4 y^{2} = 4$ by $4$ and simplify.

Divide each term in $4 y^{2} = 4$ by $4$ .

$\frac{4 y^{2}}{4} = \frac{4}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{2}}{4} = \frac{4}{4}$

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{4}{4}$

Simplify the right side.

Divide $4$ by $4$ .

$y^{2} = 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$y = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 1, - 1$

y-intercept(s) in point form.

y-intercept(s): $(0, 1), (0, - 1)$

List the intersections.

x-intercept(s): $(2, 0), (- 2, 0)$

y-intercept(s): $(0, 1), (0, - 1)$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.