Find the x and y Intercepts 4x^2+2x-3

$4 x^{2} + 2 x - 3$

Step 1

Write $4 x^{2} + 2 x - 3$ as an equation.

$y = 4 x^{2} + 2 x - 3$

Step 2

Find the x-intercepts.

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

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

Solve the equation.

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

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 2$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (4 \cdot - 3)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot - 3}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 3$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16 \cdot - 3}}{2 \cdot 4}$

Multiply $- 16$ by $- 3$ .

$x = \frac{- 2 \pm \sqrt{4 + 48}}{2 \cdot 4}$

Add $4$ and $48$ .

$x = \frac{- 2 \pm \sqrt{52}}{2 \cdot 4}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{- 2 \pm \sqrt{4 (13)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{13}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2 \sqrt{13}}{8}$

Simplify $\frac{- 2 \pm 2 \sqrt{13}}{8}$ .

$x = \frac{- 1 \pm \sqrt{13}}{4}$

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

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot - 3}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 3$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16 \cdot - 3}}{2 \cdot 4}$

Multiply $- 16$ by $- 3$ .

$x = \frac{- 2 \pm \sqrt{4 + 48}}{2 \cdot 4}$

Add $4$ and $48$ .

$x = \frac{- 2 \pm \sqrt{52}}{2 \cdot 4}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{- 2 \pm \sqrt{4 (13)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{13}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2 \sqrt{13}}{8}$

Simplify $\frac{- 2 \pm 2 \sqrt{13}}{8}$ .

$x = \frac{- 1 \pm \sqrt{13}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{- 1 + \sqrt{13}}{4}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 + \sqrt{13}}{4}$

Factor $- 1$ out of $\sqrt{13}$ .

$x = \frac{- 1 \cdot 1 - 1 (- \sqrt{13})}{4}$

Factor $- 1$ out of $- 1 (1) - 1 (- \sqrt{13})$ .

$x = \frac{- 1 (1 - \sqrt{13})}{4}$

Move the negative in front of the fraction.

$x = - \frac{1 - \sqrt{13}}{4}$

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

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot - 3}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 3$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16 \cdot - 3}}{2 \cdot 4}$

Multiply $- 16$ by $- 3$ .

$x = \frac{- 2 \pm \sqrt{4 + 48}}{2 \cdot 4}$

Add $4$ and $48$ .

$x = \frac{- 2 \pm \sqrt{52}}{2 \cdot 4}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{- 2 \pm \sqrt{4 (13)}}{2 \cdot 4}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{13}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 2 \pm 2 \sqrt{13}}{8}$

Simplify $\frac{- 2 \pm 2 \sqrt{13}}{8}$ .

$x = \frac{- 1 \pm \sqrt{13}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{- 1 - \sqrt{13}}{4}$

Rewrite $- 1$ as $- 1 (1)$ .

$x = \frac{- 1 \cdot 1 - \sqrt{13}}{4}$

Factor $- 1$ out of $- \sqrt{13}$ .

$x = \frac{- 1 \cdot 1 - (\sqrt{13})}{4}$

Factor $- 1$ out of $- 1 (1) - (\sqrt{13})$ .

$x = \frac{- 1 (1 + \sqrt{13})}{4}$

Move the negative in front of the fraction.

$x = - \frac{1 + \sqrt{13}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{13}}{4}, - \frac{1 + \sqrt{13}}{4}$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{1 - \sqrt{13}}{4}, 0), (- \frac{1 + \sqrt{13}}{4}, 0)$

Step 3

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

$y = 4 \cdot 0 + 2 (0) - 3$

Multiply $4$ by $0$ .

$y = 0 + 2 (0) - 3$

Multiply $2$ by $0$ .

$y = 0 + 0 - 3$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

Step 4

List the intersections.

x-intercept(s): $(- \frac{1 - \sqrt{13}}{4}, 0), (- \frac{1 + \sqrt{13}}{4}, 0)$

y-intercept(s): $(0, - 3)$

Step 5

Find the x and y Intercepts 4x^2+2x-3

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