Find the x and y Intercepts y=x^2 square root of 25-x^2

$y = x^{2} \sqrt{25 - x^{2}}$

Find the x-intercepts.

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

$0 = x^{2} \sqrt{25 - x^{2}}$

Solve the equation.

Rewrite the equation as $x^{2} \sqrt{25 - x^{2}} = 0$ .

$x^{2} \sqrt{25 - x^{2}} = 0$

To remove the radical on the left side of the equation, square both sides of the equation.

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{25 - x^{2}}$ as ${(25 - x^{2})}^{\frac{1}{2}}$ .

${(x^{2} {(25 - x^{2})}^{\frac{1}{2}})}^{2} = 0^{2}$

Simplify the left side.

Simplify ${(x^{2} {(25 - x^{2})}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $x^{2} {(25 - x^{2})}^{\frac{1}{2}}$ .

${(x^{2})}^{2} {({(25 - x^{2})}^{\frac{1}{2}})}^{2} = 0^{2}$

Multiply the exponents in ${(x^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2 \cdot 2} {({(25 - x^{2})}^{\frac{1}{2}})}^{2} = 0^{2}$

Multiply $2$ by $2$ .

$x^{4} {({(25 - x^{2})}^{\frac{1}{2}})}^{2} = 0^{2}$

Multiply the exponents in ${({(25 - x^{2})}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{4} {(25 - x^{2})}^{\frac{1}{2} \cdot 2} = 0^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{4} {(25 - x^{2})}^{\frac{1}{2} \cdot 2} = 0^{2}$

Rewrite the expression.

$x^{4} {(25 - x^{2})}^{1} = 0^{2}$

Simplify.

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

Simplify by multiplying through.

Apply the distributive property.

$x^{4} \cdot 25 + x^{4} (- x^{2}) = 0^{2}$

Reorder.

Move $25$ to the left of $x^{4}$ .

$25 \cdot x^{4} + x^{4} (- x^{2}) = 0^{2}$

Rewrite using the commutative property of multiplication.

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

Multiply $x^{4}$ by $x^{2}$ by adding the exponents.

Move $x^{2}$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $2$ and $4$ .

$25 x^{4} - x^{6} = 0^{2}$

Simplify the right side.

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

$25 x^{4} - x^{6} = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x^{4}$ out of $25 x^{4} - x^{6}$ .

Factor $x^{4}$ out of $25 x^{4}$ .

$x^{4} \cdot 25 - x^{6} = 0$

Factor $x^{4}$ out of $- x^{6}$ .

$x^{4} \cdot 25 + x^{4} (- x^{2}) = 0$

Factor $x^{4}$ out of $x^{4} \cdot 25 + x^{4} (- x^{2})$ .

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

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

$x^{4} (5^{2} - x^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$x^{4} ((5 + x) (5 - x)) = 0$

Remove unnecessary parentheses.

$x^{4} (5 + x) (5 - x) = 0$

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

$x^{4} = 0$

$5 + x = 0$

$5 - x = 0$

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

Set $x^{4}$ equal to $0$ .

$x^{4} = 0$

Solve $x^{4} = 0$ for $x$ .

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{0}$

Simplify $\pm \sqrt[4]{0}$ .

Rewrite $0$ as $0^{4}$ .

$x = \pm \sqrt[4]{0^{4}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $5 + x$ equal to $0$ and solve for $x$ .

Set $5 + x$ equal to $0$ .

$5 + x = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $5 - x$ equal to $0$ and solve for $x$ .

Set $5 - x$ equal to $0$ .

$5 - x = 0$

Solve $5 - x = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

The final solution is all the values that make $x^{4} (5 + x) (5 - x) = 0$ true.

$x = 0, - 5, 5$

x-intercept(s) in point form.

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

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

$y = 0^{2} \sqrt{25 - 0^{2}}$

Remove parentheses.

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

Simplify ${(0)}^{2} \sqrt{25 - {(0)}^{2}}$ .

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

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

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

$y = 0 \sqrt{25 - 0}$

Multiply $- 1$ by $0$ .

$y = 0 \sqrt{25 + 0}$

Add $25$ and $0$ .

$y = 0 \sqrt{25}$

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

$y = 0 \sqrt{5^{2}}$

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

$y = 0 \cdot 5$

Multiply $0$ by $5$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

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

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

Find the x and y Intercepts y=x^2 square root of 25-x^2

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