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

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

Step 1

Find the x-intercepts.

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

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

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

Simplify each term.

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

$x - 0 = - 4$

Multiply $- 1$ by $0$ .

$x + 0 = - 4$

Add $x$ and $0$ .

$x = - 4$

x-intercept(s) in point form.

x-intercept(s): $(- 4, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Subtract $y^{2}$ from $0$ .

$- y^{2} = - 4$

Divide each term in $- y^{2} = - 4$ by $- 1$ and simplify.

Divide each term in $- y^{2} = - 4$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{- 4}{- 1}$

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

$y^{2} = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$y^{2} = 4$

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

$y = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$y = \pm 2$

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.

$y = 2$

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

$y = - 2$

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

$y = 2, - 2$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(- 4, 0)$

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

Step 4

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

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