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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{3 \pm \sqrt{9 - 24}}{2 \cdot 2}$

Subtract $24$ from $9$ .

$x = \frac{3 \pm \sqrt{- 15}}{2 \cdot 2}$

Rewrite $- 15$ as $- 1 (15)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 15}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (15)}$ as $\sqrt{- 1} \cdot \sqrt{15}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{15}}{2 \cdot 2}$

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

$x = \frac{3 \pm i \sqrt{15}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm i \sqrt{15}}{4}$

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

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{3 \pm \sqrt{9 - 24}}{2 \cdot 2}$

Subtract $24$ from $9$ .

$x = \frac{3 \pm \sqrt{- 15}}{2 \cdot 2}$

Rewrite $- 15$ as $- 1 (15)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 15}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (15)}$ as $\sqrt{- 1} \cdot \sqrt{15}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{15}}{2 \cdot 2}$

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

$x = \frac{3 \pm i \sqrt{15}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm i \sqrt{15}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{3 + i \sqrt{15}}{4}$

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

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 3$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{3 \pm \sqrt{9 - 24}}{2 \cdot 2}$

Subtract $24$ from $9$ .

$x = \frac{3 \pm \sqrt{- 15}}{2 \cdot 2}$

Rewrite $- 15$ as $- 1 (15)$ .

$x = \frac{3 \pm \sqrt{- 1 \cdot 15}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (15)}$ as $\sqrt{- 1} \cdot \sqrt{15}$ .

$x = \frac{3 \pm \sqrt{- 1} \cdot \sqrt{15}}{2 \cdot 2}$

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

$x = \frac{3 \pm i \sqrt{15}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm i \sqrt{15}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{3 - i \sqrt{15}}{4}$

The final answer is the combination of both solutions.

$x = \frac{3 + i \sqrt{15}}{4}, \frac{3 - i \sqrt{15}}{4}$

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

x-intercept(s): $None$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

$y = 2 \cdot 0 - 3 \cdot 0 + 3$

Multiply $2$ by $0$ .

$y = 0 - 3 \cdot 0 + 3$

Multiply $- 3$ by $0$ .

$y = 0 + 0 + 3$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 3$

Add $0$ and $3$ .

$y = 3$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $None$

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

Step 4

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

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