Find the x and y Intercepts f(x)=-4x(x^2+1)(2x+1)^2(x+6)^2

$f (x) = - 4 x (x^{2} + 1) {(2 x + 1)}^{2} {(x + 6)}^{2}$

Step 1

Find the x-intercepts.

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

$0 = - 4 x (x^{2} + 1) {(2 x + 1)}^{2} {(x + 6)}^{2}$

Solve the equation.

Rewrite the equation as $- 4 x (x^{2} + 1) {(2 x + 1)}^{2} {(x + 6)}^{2} = 0$ .

$- 4 x (x^{2} + 1) {(2 x + 1)}^{2} {(x + 6)}^{2} = 0$

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

$x = 0$

$x^{2} + 1 = 0$

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

${(x + 6)}^{2} = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

Set ${(2 x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(2 x + 1)}^{2}$ equal to $0$ .

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

Solve ${(2 x + 1)}^{2} = 0$ for $x$ .

Set the $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set ${(x + 6)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 6)}^{2}$ equal to $0$ .

${(x + 6)}^{2} = 0$

Solve ${(x + 6)}^{2} = 0$ for $x$ .

Set the $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- 4 x (x^{2} + 1) {(2 x + 1)}^{2} {(x + 6)}^{2} = 0$ true.

$x = 0, i, - i, - \frac{1}{2}, - 6$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (- \frac{1}{2}, 0), (- 6, 0)$

Step 2

Find the y-intercepts.

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

$y = - 4 \cdot 0 ({(0)}^{2} + 1) {(2 (0) + 1)}^{2} {((0) + 6)}^{2}$

Solve the equation.

Remove parentheses.

$y = - 4 \cdot (0 (0^{2} + 1) {(2 (0) + 1)}^{2} {((0) + 6)}^{2})$

Remove parentheses.

$y = - 4 \cdot (0 (0^{2} + 1) {(2 (0) + 1)}^{2} {(0 + 6)}^{2})$

Remove parentheses.

$y = - 4 \cdot (0 ({(0)}^{2} + 1) {(2 (0) + 1)}^{2} {((0) + 6)}^{2})$

Simplify $- 4 \cdot (0 ({(0)}^{2} + 1) {(2 (0) + 1)}^{2} {((0) + 6)}^{2})$ .

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

$y = - 4 \cdot (0 (0 + 1) {(2 (0) + 1)}^{2} {((0) + 6)}^{2})$

Add $0$ and $1$ .

$y = - 4 \cdot (0 \cdot 1 {(2 (0) + 1)}^{2} {((0) + 6)}^{2})$

Multiply $0$ by $1$ .

$y = - 4 \cdot (0 {(2 (0) + 1)}^{2} {((0) + 6)}^{2})$

Multiply $2$ by $0$ .

$y = - 4 \cdot (0 {(0 + 1)}^{2} {((0) + 6)}^{2})$

Add $0$ and $1$ .

$y = - 4 \cdot (0 \cdot 1^{2} {((0) + 6)}^{2})$

One to any power is one.

$y = - 4 \cdot (0 \cdot 1 {((0) + 6)}^{2})$

Multiply $0$ by $1$ .

$y = - 4 \cdot (0 {((0) + 6)}^{2})$

Add $0$ and $6$ .

$y = - 4 \cdot (0 \cdot 6^{2})$

Raise $6$ to the power of $2$ .

$y = - 4 \cdot (0 \cdot 36)$

Multiply $- 4 (0 \cdot 36)$ .

Multiply $0$ by $36$ .

$y = - 4 \cdot 0$

Multiply $- 4$ by $0$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(0, 0), (- \frac{1}{2}, 0), (- 6, 0)$

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

Step 4

Find the x and y Intercepts f(x)=-4x(x^2+1)(2x+1)^2(x+6)^2

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