Find the x and y Intercepts f(x) = fourth root of 3x^2+1

$f (x) = \sqrt[4]{3 x^{2} + 1}$

Find the x-intercepts.

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

$0 = \sqrt[4]{3 x^{2} + 1}$

Solve the equation.

Rewrite the equation as $\sqrt[4]{3 x^{2} + 1} = 0$ .

$\sqrt[4]{3 x^{2} + 1} = 0$

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $4$ .

${\sqrt[4]{3 x^{2} + 1}}^{4} = 0^{4}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{3 x^{2} + 1}$ as ${(3 x^{2} + 1)}^{\frac{1}{4}}$ .

${({(3 x^{2} + 1)}^{\frac{1}{4}})}^{4} = 0^{4}$

Simplify the left side.

Simplify ${({(3 x^{2} + 1)}^{\frac{1}{4}})}^{4}$ .

Multiply the exponents in ${({(3 x^{2} + 1)}^{\frac{1}{4}})}^{4}$ .

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

${(3 x^{2} + 1)}^{\frac{1}{4} \cdot 4} = 0^{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(3 x^{2} + 1)}^{\frac{1}{4} \cdot 4} = 0^{4}$

Rewrite the expression.

${(3 x^{2} + 1)}^{1} = 0^{4}$

Simplify.

$3 x^{2} + 1 = 0^{4}$

Simplify the right side.

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

$3 x^{2} + 1 = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$3 x^{2} = - 1$

Divide each term in $3 x^{2} = - 1$ by $3$ and simplify.

Divide each term in $3 x^{2} = - 1$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{- 1}{3}$

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

$x^{2} = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} = - \frac{1}{3}$

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

$x = \pm \sqrt{- \frac{1}{3}}$

Simplify $\pm \sqrt{- \frac{1}{3}}$ .

Rewrite $- \frac{1}{3}$ as $i^{2} \frac{1^{2}}{3}$ .

Rewrite $- 1$ as $i^{2}$ .

$x = \pm \sqrt{i^{2} \frac{1}{3}}$

Rewrite $1$ as $1^{2}$ .

$x = \pm \sqrt{i^{2} \frac{1^{2}}{3}}$

Pull terms out from under the radical.

$x = \pm i \sqrt{\frac{1^{2}}{3}}$

One to any power is one.

$x = \pm i \sqrt{\frac{1}{3}}$

Rewrite $\sqrt{\frac{1}{3}}$ as $\frac{\sqrt{1}}{\sqrt{3}}$ .

$x = \pm i \frac{\sqrt{1}}{\sqrt{3}}$

Any root of $1$ is $1$ .

$x = \pm i \frac{1}{\sqrt{3}}$

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$x = \pm i (\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$x = \pm i \frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm i \frac{\sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm i \frac{\sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

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

$x = \pm i \frac{\sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$x = \pm i \frac{\sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$x = \pm i \frac{\sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

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

$x = \pm i \frac{\sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$x = \pm i \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \pm i \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$x = \pm i \frac{\sqrt{3}}{3^{1}}$

Evaluate the exponent.

$x = \pm i \frac{\sqrt{3}}{3}$

Combine $i$ and $\frac{\sqrt{3}}{3}$ .

$x = \pm \frac{i \sqrt{3}}{3}$

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.

$x = \frac{i \sqrt{3}}{3}$

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

$x = - \frac{i \sqrt{3}}{3}$

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

$x = \frac{i \sqrt{3}}{3}, - \frac{i \sqrt{3}}{3}$

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

x-intercept(s): $None$

Find the y-intercepts.

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

$y = \sqrt[4]{3 {(0)}^{2} + 1}$

Simplify $\sqrt[4]{3 {(0)}^{2} + 1}$ .

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

$y = \sqrt[4]{3 \cdot 0 + 1}$

Multiply $3$ by $0$ .

$y = \sqrt[4]{0 + 1}$

Add $0$ and $1$ .

$y = \sqrt[4]{1}$

Any root of $1$ is $1$ .

$y = 1$

y-intercept(s) in point form.

y-intercept(s): $(0, 1)$

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, 1)$

Find the x and y Intercepts f(x) = fourth root of 3x^2+1

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