Find the x and y Intercepts f(x)=-x^4+25x^2

$f (x) = - x^{4} + 25 x^{2}$

Step 1

Find the x-intercepts.

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

$0 = - x^{4} + 25 x^{2}$

Solve the equation.

Rewrite the equation as $- x^{4} + 25 x^{2} = 0$ .

$- x^{4} + 25 x^{2} = 0$

Factor the left side of the equation.

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

$- {(x^{2})}^{2} + 25 x^{2} = 0$

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

$- u^{2} + 25 u = 0$

Factor $u$ out of $- u^{2} + 25 u$ .

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

$u (- u) + 25 u = 0$

Factor $u$ out of $25 u$ .

$u (- u) + u \cdot 25 = 0$

Factor $u$ out of $u (- u) + u \cdot 25$ .

$u (- u + 25) = 0$

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

$x^{2} (- x^{2} + 25) = 0$

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

$x^{2} = 0$

$- x^{2} + 25 = 0$

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 specified 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 $- x^{2} + 25$ equal to $0$ and solve for $x$ .

Set $- x^{2} + 25$ equal to $0$ .

$- x^{2} + 25 = 0$

Solve $- x^{2} + 25 = 0$ for $x$ .

Subtract $25$ from both sides of the equation.

$- x^{2} = - 25$

Divide each term in $- x^{2} = - 25$ by $- 1$ and simplify.

Divide each term in $- x^{2} = - 25$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 25}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 25}{- 1}$

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

$x^{2} = \frac{- 25}{- 1}$

Simplify the right side.

Divide $- 25$ by $- 1$ .

$x^{2} = 25$

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

The final solution is all the values that make $x^{2} (- x^{2} + 25) = 0$ true.

$x = 0, 5, - 5$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (5, 0), (- 5, 0)$

Step 2

Find the y-intercepts.

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

$y = - {(0)}^{4} + 25 {(0)}^{2}$

Solve the equation.

Remove parentheses.

$y = - 0^{4} + 25 {(0)}^{2}$

Remove parentheses.

$y = - 0^{4} + 25 \cdot 0^{2}$

Remove parentheses.

$y = - {(0)}^{4} + 25 {(0)}^{2}$

Simplify $- {(0)}^{4} + 25 {(0)}^{2}$ .

Simplify each term.

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

$y = - 0 + 25 {(0)}^{2}$

Multiply $- 1$ by $0$ .

$y = 0 + 25 {(0)}^{2}$

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

$y = 0 + 25 \cdot 0$

Multiply $25$ by $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(0, 0), (5, 0), (- 5, 0)$

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

Step 4

Find the x and y Intercepts f(x)=-x^4+25x^2

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