Find the x and y Intercepts f(x)=(2x^2+x+3)/(2x^2+3x+1)

$f (x) = \frac{2 x^{2} + x + 3}{2 x^{2} + 3 x + 1}$

Find the x-intercepts.

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

$0 = \frac{2 x^{2} + x + 3}{2 x^{2} + 3 x + 1}$

Solve the equation.

Set the numerator equal to zero.

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

Solve the equation for $x$ .

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 $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (2 \cdot 3)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{- 1 \pm \sqrt{1 - 24}}{2 \cdot 2}$

Subtract $24$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 23}}{2 \cdot 2}$

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 2}$

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

$x = \frac{- 1 \pm i \sqrt{23}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm i \sqrt{23}}{4}$

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

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{- 1 \pm \sqrt{1 - 24}}{2 \cdot 2}$

Subtract $24$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 23}}{2 \cdot 2}$

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 2}$

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

$x = \frac{- 1 \pm i \sqrt{23}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm i \sqrt{23}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{- 1 + i \sqrt{23}}{4}$

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

$x = \frac{- 1 \cdot 1 + i \sqrt{23}}{4}$

Factor $- 1$ out of $i \sqrt{23}$ .

$x = \frac{- 1 \cdot 1 - (- i \sqrt{23})}{4}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{23})$ .

$x = \frac{- 1 (1 - i \sqrt{23})}{4}$

Move the negative in front of the fraction.

$x = - \frac{1 - i \sqrt{23}}{4}$

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

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot 3}}{2 \cdot 2}$

Multiply $- 8$ by $3$ .

$x = \frac{- 1 \pm \sqrt{1 - 24}}{2 \cdot 2}$

Subtract $24$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 23}}{2 \cdot 2}$

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 23}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot 2}$

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

$x = \frac{- 1 \pm i \sqrt{23}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm i \sqrt{23}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{- 1 - i \sqrt{23}}{4}$

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

$x = \frac{- 1 \cdot 1 - i \sqrt{23}}{4}$

Factor $- 1$ out of $- i \sqrt{23}$ .

$x = \frac{- 1 \cdot 1 - (i \sqrt{23})}{4}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{23})$ .

$x = \frac{- 1 (1 + i \sqrt{23})}{4}$

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{23}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{23}}{4}, - \frac{1 + i \sqrt{23}}{4}$

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 $y$ .

$y = \frac{2 {(0)}^{2} + 0 + 3}{2 {(0)}^{2} + 3 (0) + 1}$

Solve the equation.

Remove parentheses.

$y = \frac{2 {(0)}^{2} + 0 + 3}{2 {(0)}^{2} + 3 (0) + 1}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} + 0 + 3}{2 {(0)}^{2} + 3 (0) + 1}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} + 0 + 3}{2 \cdot 0^{2} + 3 (0) + 1}$

Remove parentheses.

$y = \frac{2 {(0)}^{2} + 0 + 3}{2 {(0)}^{2} + 3 (0) + 1}$

Simplify $\frac{2 {(0)}^{2} + 0 + 3}{2 {(0)}^{2} + 3 (0) + 1}$ .

Simplify the numerator.

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

$y = \frac{2 \cdot 0 + 0 + 3}{2 \cdot 0^{2} + 3 (0) + 1}$

Multiply $2$ by $0$ .

$y = \frac{0 + 0 + 3}{2 \cdot 0^{2} + 3 (0) + 1}$

Add $0$ and $0$ .

$y = \frac{0 + 3}{2 \cdot 0^{2} + 3 (0) + 1}$

Add $0$ and $3$ .

$y = \frac{3}{2 \cdot 0^{2} + 3 (0) + 1}$

Simplify the denominator.

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

$y = \frac{3}{2 \cdot 0 + 3 (0) + 1}$

Multiply $2$ by $0$ .

$y = \frac{3}{0 + 3 (0) + 1}$

Multiply $3$ by $0$ .

$y = \frac{3}{0 + 0 + 1}$

Add $0$ and $0$ .

$y = \frac{3}{0 + 1}$

Add $0$ and $1$ .

$y = \frac{3}{1}$

Divide $3$ by $1$ .

$y = 3$

y-intercept(s) in point form.

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

List the intersections.

x-intercept(s): $None$

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

Find the x and y Intercepts f(x)=(2x^2+x+3)/(2x^2+3x+1)

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