Find the x and y Intercepts F(x)=(2x^2)/(x^2+4)

$F (x) = \frac{2 x^{2}}{x^{2} + 4}$

Find the x-intercepts.

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

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

Solve the equation.

Set the numerator equal to zero.

$2 x^{2} = 0$

Solve the equation for $x$ .

Divide each term in $2 x^{2} = 0$ by $2$ and simplify.

Divide each term in $2 x^{2} = 0$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{0}{2}$

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

$x^{2} = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

x-intercept(s) in point form.

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

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

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

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

Simplify the denominator.

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

$y = \frac{2 \cdot 0}{0 + 4}$

Add $0$ and $4$ .

$y = \frac{2 \cdot 0}{4}$

Simplify the expression.

Multiply $2$ by $0$ .

$y = \frac{0}{4}$

Divide $0$ by $4$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

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

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

Find the x and y Intercepts F(x)=(2x^2)/(x^2+4)

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