Find the x and y Intercepts f(x)=-2x^2(2x-1)^3(4x+3)

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

Rewrite the equation as $- 2 x^{2} {(2 x - 1)}^{3} (4 x + 3) = 0$ .

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

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

$x^{2} = 0$

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

$4 x + 3 = 0$

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 square 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 ${(2 x - 1)}^{3}$ equal to $0$ and solve for $x$ .

Set ${(2 x - 1)}^{3}$ equal to $0$ .

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

Solve ${(2 x - 1)}^{3} = 0$ for $x$ .

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

$2 x - 1 = 0$

Solve for $x$ .

Add $1$ to 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 $1$ .

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

Set $4 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Solve $4 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Divide each term in $4 x = - 3$ by $4$ and simplify.

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

The final solution is all the values that make $- 2 x^{2} {(2 x - 1)}^{3} (4 x + 3) = 0$ true.

$x = 0, \frac{1}{2}, - \frac{3}{4}$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (\frac{1}{2}, 0), (- \frac{3}{4}, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

$y = - 2 \cdot (0^{2} {(2 (0) - 1)}^{3} (4 (0) + 3))$

Remove parentheses.

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

Simplify $- 2 {(0)}^{2} {(2 (0) - 1)}^{3} (4 (0) + 3)$ .

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

$y = - 2 \cdot 0 {(2 (0) - 1)}^{3} (4 (0) + 3)$

Multiply $- 2$ by $0$ .

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

Multiply $2$ by $0$ .

$y = 0 {(0 - 1)}^{3} (4 (0) + 3)$

Subtract $1$ from $0$ .

$y = 0 {(- 1)}^{3} (4 (0) + 3)$

Raise $- 1$ to the power of $3$ .

$y = 0 \cdot - 1 (4 (0) + 3)$

Multiply $0$ by $- 1$ .

$y = 0 (4 (0) + 3)$

Multiply $4$ by $0$ .

$y = 0 (0 + 3)$

Add $0$ and $3$ .

$y = 0 \cdot 3$

Multiply $0$ by $3$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(0, 0), (\frac{1}{2}, 0), (- \frac{3}{4}, 0)$

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

Step 4

Find the x and y Intercepts f(x)=-2x^2(2x-1)^3(4x+3)

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