Find the x and y Intercepts f(x)=-x^4+4x^2

$f (x) = - x^{4} + 4 x^{2}$

Find the x-intercepts.

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

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

Solve the equation.

Rewrite the equation as $- x^{4} + 4 x^{2} = 0$ .

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

Factor the left side of the equation.

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

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

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

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

Factor $u$ out of $- u^{2} + 4 u$ .

Factor $u$ out of $- u^{2}$ .

$u (- u) + 4 u = 0$

Factor $u$ out of $4 u$ .

$u (- u) + u \cdot 4 = 0$

Factor $u$ out of $u (- u) + u \cdot 4$ .

$u (- u + 4) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} = 0$

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

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

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 $0$ .

$x = 0$

Set $- x^{2} + 4$ equal to $0$ and solve for $x$ .

Set $- x^{2} + 4$ equal to $0$ .

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

Solve $- x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- x^{2} = - 4$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

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

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

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

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.

$x = 2, - 2$

The final solution is all the values that make $x^{2} (- x^{2} + 4) = 0$ true.

$x = 0, 2, - 2$

x-intercept(s) in point form.

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

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

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

Multiply $- 1$ by $0$ .

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

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

$y = 0 + 4 \cdot 0$

Multiply $4$ by $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

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

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

Find the x and y Intercepts f(x)=-x^4+4x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.