Find the x and y Intercepts -3x^2+4x-7

$- 3 x^{2} + 4 x - 7$

Step 1

Write $- 3 x^{2} + 4 x - 7$ as an equation.

$y = - 3 x^{2} + 4 x - 7$

Step 2

Find the x-intercepts.

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

$0 = - 3 x^{2} + 4 x - 7$

Solve the equation.

Rewrite the equation as $- 3 x^{2} + 4 x - 7 = 0$ .

$- 3 x^{2} + 4 x - 7 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 3$ , $b = 4$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (- 3 \cdot - 7)}}{2 \cdot - 3}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3 \cdot - 7}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 7$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12 \cdot - 7}}{2 \cdot - 3}$

Multiply $12$ by $- 7$ .

$x = \frac{- 4 \pm \sqrt{16 - 84}}{2 \cdot - 3}$

Subtract $84$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 68}}{2 \cdot - 3}$

Rewrite $- 68$ as $- 1 (68)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 68}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1 (68)}$ as $\sqrt{- 1} \cdot \sqrt{68}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{68}}{2 \cdot - 3}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{68}}{2 \cdot - 3}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (17)}}{2 \cdot - 3}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 17}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{17})}{2 \cdot - 3}$

Move $2$ to the left of $i$ .

$x = \frac{- 4 \pm 2 i \sqrt{17}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 4 \pm 2 i \sqrt{17}}{- 6}$

Simplify $\frac{- 4 \pm 2 i \sqrt{17}}{- 6}$ .

$x = \frac{2 \pm i \sqrt{17}}{3}$

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

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3 \cdot - 7}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 7$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12 \cdot - 7}}{2 \cdot - 3}$

Multiply $12$ by $- 7$ .

$x = \frac{- 4 \pm \sqrt{16 - 84}}{2 \cdot - 3}$

Subtract $84$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 68}}{2 \cdot - 3}$

Rewrite $- 68$ as $- 1 (68)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 68}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1 (68)}$ as $\sqrt{- 1} \cdot \sqrt{68}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{68}}{2 \cdot - 3}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{68}}{2 \cdot - 3}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (17)}}{2 \cdot - 3}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 17}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{17})}{2 \cdot - 3}$

Move $2$ to the left of $i$ .

$x = \frac{- 4 \pm 2 i \sqrt{17}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 4 \pm 2 i \sqrt{17}}{- 6}$

Simplify $\frac{- 4 \pm 2 i \sqrt{17}}{- 6}$ .

$x = \frac{2 \pm i \sqrt{17}}{3}$

Change the $\pm$ to $+$ .

$x = \frac{2 + i \sqrt{17}}{3}$

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

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3 \cdot - 7}}{2 \cdot - 3}$

Multiply $- 4 \cdot - 3 \cdot - 7$ .

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12 \cdot - 7}}{2 \cdot - 3}$

Multiply $12$ by $- 7$ .

$x = \frac{- 4 \pm \sqrt{16 - 84}}{2 \cdot - 3}$

Subtract $84$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 68}}{2 \cdot - 3}$

Rewrite $- 68$ as $- 1 (68)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 68}}{2 \cdot - 3}$

Rewrite $\sqrt{- 1 (68)}$ as $\sqrt{- 1} \cdot \sqrt{68}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{68}}{2 \cdot - 3}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{68}}{2 \cdot - 3}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4 (17)}}{2 \cdot - 3}$

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

$x = \frac{- 4 \pm i \cdot \sqrt{2^{2} \cdot 17}}{2 \cdot - 3}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (2 \sqrt{17})}{2 \cdot - 3}$

Move $2$ to the left of $i$ .

$x = \frac{- 4 \pm 2 i \sqrt{17}}{2 \cdot - 3}$

Multiply $2$ by $- 3$ .

$x = \frac{- 4 \pm 2 i \sqrt{17}}{- 6}$

Simplify $\frac{- 4 \pm 2 i \sqrt{17}}{- 6}$ .

$x = \frac{2 \pm i \sqrt{17}}{3}$

Change the $\pm$ to $-$ .

$x = \frac{2 - i \sqrt{17}}{3}$

The final answer is the combination of both solutions.

$x = \frac{2 + i \sqrt{17}}{3}, \frac{2 - i \sqrt{17}}{3}$

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

x-intercept(s): $None$

Step 3

Find the y-intercepts.

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

$y = - 3 {(0)}^{2} + 4 (0) - 7$

Solve the equation.

Remove parentheses.

$y = - 3 \cdot 0^{2} + 4 (0) - 7$

Remove parentheses.

$y = - 3 {(0)}^{2} + 4 (0) - 7$

Simplify $- 3 {(0)}^{2} + 4 (0) - 7$ .

Simplify each term.

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

$y = - 3 \cdot 0 + 4 (0) - 7$

Multiply $- 3$ by $0$ .

$y = 0 + 4 (0) - 7$

Multiply $4$ by $0$ .

$y = 0 + 0 - 7$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 7$

Subtract $7$ from $0$ .

$y = - 7$

y-intercept(s) in point form.

y-intercept(s): $(0, - 7)$

Step 4

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, - 7)$

Step 5

Find the x and y Intercepts -3x^2+4x-7

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