Find the x and y Intercepts F(x)=x^4-10x^2+9

$F (x) = x^{4} - 10 x^{2} + 9$

Find the x-intercepts.

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

$0 = x^{4} - 10 x^{2} + 9$

Solve the equation.

Rewrite the equation as $x^{4} - 10 x^{2} + 9 = 0$ .

$x^{4} - 10 x^{2} + 9 = 0$

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 10 u + 9 = 0$

$u = x^{2}$

Factor $u^{2} - 10 u + 9$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $9$ and whose sum is $- 10$ .

$- 9, - 1$

Write the factored form using these integers.

$(u - 9) (u - 1) = 0$

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

$u - 9 = 0$

$u - 1 = 0$

Set $u - 9$ equal to $0$ and solve for $u$ .

Set $u - 9$ equal to $0$ .

$u - 9 = 0$

Add $9$ to both sides of the equation.

$u = 9$

Set $u - 1$ equal to $0$ and solve for $u$ .

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

The final solution is all the values that make $(u - 9) (u - 1) = 0$ true.

$u = 9, 1$

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 9$

${(x^{2})}^{1} = 1$

Solve the first equation for $x$ .

$x^{2} = 9$

Solve the equation for $x$ .

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

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

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

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

$x = \pm 3$

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.

$x = 3$

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

$x = - 3$

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

$x = 3, - 3$

Solve the second equation for $x$ .

${(x^{2})}^{1} = 1$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

The solution to $x^{4} - 10 x^{2} + 9 = 0$ is $x = 3, - 3, 1, - 1$ .

$x = 3, - 3, 1, - 1$

x-intercept(s) in point form.

x-intercept(s): $(3, 0), (- 3, 0), (1, 0), (- 1, 0)$

Find the y-intercepts.

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

$y = {(0)}^{4} - 10 {(0)}^{2} + 9$

Solve the equation.

Remove parentheses.

$y = 0^{4} - 10 {(0)}^{2} + 9$

Remove parentheses.

$y = 0^{4} - 10 \cdot 0^{2} + 9$

Remove parentheses.

$y = {(0)}^{4} - 10 {(0)}^{2} + 9$

Simplify ${(0)}^{4} - 10 {(0)}^{2} + 9$ .

Simplify each term.

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

$y = 0 - 10 {(0)}^{2} + 9$

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

$y = 0 - 10 \cdot 0 + 9$

Multiply $- 10$ by $0$ .

$y = 0 + 0 + 9$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 9$

Add $0$ and $9$ .

$y = 9$

y-intercept(s) in point form.

y-intercept(s): $(0, 9)$

List the intersections.

x-intercept(s): $(3, 0), (- 3, 0), (1, 0), (- 1, 0)$

y-intercept(s): $(0, 9)$

Find the x and y Intercepts F(x)=x^4-10x^2+9

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