Find the x and y Intercepts y=4x^2-x+8

$y = 4 x^{2} - x + 8$

Find the x-intercepts.

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

$0 = 4 x^{2} - x + 8$

Solve the equation.

Rewrite the equation as $4 x^{2} - x + 8 = 0$ .

$4 x^{2} - x + 8 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = - 1$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (4 \cdot 8)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 4 \cdot 8}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{1 \pm \sqrt{1 - 16 \cdot 8}}{2 \cdot 4}$

Multiply $- 16$ by $8$ .

$x = \frac{1 \pm \sqrt{1 - 128}}{2 \cdot 4}$

Subtract $128$ from $1$ .

$x = \frac{1 \pm \sqrt{- 127}}{2 \cdot 4}$

Rewrite $- 127$ as $- 1 (127)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 127}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (127)}$ as $\sqrt{- 1} \cdot \sqrt{127}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{127}}{2 \cdot 4}$

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

$x = \frac{1 \pm i \sqrt{127}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{1 \pm i \sqrt{127}}{8}$

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

Simplify the numerator.

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 4 \cdot 8}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{1 \pm \sqrt{1 - 16 \cdot 8}}{2 \cdot 4}$

Multiply $- 16$ by $8$ .

$x = \frac{1 \pm \sqrt{1 - 128}}{2 \cdot 4}$

Subtract $128$ from $1$ .

$x = \frac{1 \pm \sqrt{- 127}}{2 \cdot 4}$

Rewrite $- 127$ as $- 1 (127)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 127}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (127)}$ as $\sqrt{- 1} \cdot \sqrt{127}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{127}}{2 \cdot 4}$

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

$x = \frac{1 \pm i \sqrt{127}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{1 \pm i \sqrt{127}}{8}$

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{127}}{8}$

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

Simplify the numerator.

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 4 \cdot 8}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{1 \pm \sqrt{1 - 16 \cdot 8}}{2 \cdot 4}$

Multiply $- 16$ by $8$ .

$x = \frac{1 \pm \sqrt{1 - 128}}{2 \cdot 4}$

Subtract $128$ from $1$ .

$x = \frac{1 \pm \sqrt{- 127}}{2 \cdot 4}$

Rewrite $- 127$ as $- 1 (127)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 127}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (127)}$ as $\sqrt{- 1} \cdot \sqrt{127}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{127}}{2 \cdot 4}$

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

$x = \frac{1 \pm i \sqrt{127}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{1 \pm i \sqrt{127}}{8}$

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{127}}{8}$

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{127}}{8}, \frac{1 - i \sqrt{127}}{8}$

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

x-intercept(s): $None$

Find the y-intercepts.

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

$y = 4 {(0)}^{2} - (0) + 8$

Solve the equation.

Remove parentheses.

$y = 4 \cdot 0^{2} - (0) + 8$

Remove parentheses.

$y = 4 {(0)}^{2} - (0) + 8$

Simplify $4 {(0)}^{2} - (0) + 8$ .

Simplify each term.

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

$y = 4 \cdot 0 - (0) + 8$

Multiply $4$ by $0$ .

$y = 0 - (0) + 8$

Multiply $- 1$ by $0$ .

$y = 0 + 0 + 8$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 8$

Add $0$ and $8$ .

$y = 8$

y-intercept(s) in point form.

y-intercept(s): $(0, 8)$

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, 8)$

Find the x and y Intercepts y=4x^2-x+8

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