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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

Factor $2$ out of $4 x^{2} - 2 x - 4$ .

Factor $2$ out of $4 x^{2}$ .

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

Factor $2$ out of $- 2 x$ .

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

Factor $2$ out of $- 4$ .

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

Factor $2$ out of $2 (2 x^{2}) + 2 (- x)$ .

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

Factor $2$ out of $2 (2 x^{2} - x) + 2 (- 2)$ .

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

Divide each term in $2 (2 x^{2} - x - 2) = 0$ by $2$ and simplify.

Divide each term in $2 (2 x^{2} - x - 2) = 0$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

$2 x^{2} - x - 2 = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$2 x^{2} - x - 2 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 1$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (2 \cdot - 2)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 2$ .

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot - 2}}{2 \cdot 2}$

Multiply $- 8$ by $- 2$ .

$x = \frac{1 \pm \sqrt{1 + 16}}{2 \cdot 2}$

Add $1$ and $16$ .

$x = \frac{1 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm \sqrt{17}}{4}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 2$ .

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot - 2}}{2 \cdot 2}$

Multiply $- 8$ by $- 2$ .

$x = \frac{1 \pm \sqrt{1 + 16}}{2 \cdot 2}$

Add $1$ and $16$ .

$x = \frac{1 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm \sqrt{17}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{1 + \sqrt{17}}{4}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 2$ .

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot - 2}}{2 \cdot 2}$

Multiply $- 8$ by $- 2$ .

$x = \frac{1 \pm \sqrt{1 + 16}}{2 \cdot 2}$

Add $1$ and $16$ .

$x = \frac{1 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm \sqrt{17}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{1 - \sqrt{17}}{4}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{17}}{4}, \frac{1 - \sqrt{17}}{4}$

x-intercept(s) in point form.

x-intercept(s): $(\frac{1 + \sqrt{17}}{4}, 0), (\frac{1 - \sqrt{17}}{4}, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

$y = 4 \cdot 0 - 2 \cdot 0 - 4$

Multiply $4$ by $0$ .

$y = 0 - 2 \cdot 0 - 4$

Multiply $- 2$ by $0$ .

$y = 0 + 0 - 4$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 4$

Subtract $4$ from $0$ .

$y = - 4$

y-intercept(s) in point form.

y-intercept(s): $(0, - 4)$

Step 3

List the intersections.

x-intercept(s): $(\frac{1 + \sqrt{17}}{4}, 0), (\frac{1 - \sqrt{17}}{4}, 0)$

y-intercept(s): $(0, - 4)$

Step 4

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

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