Find the x and y Intercepts y=(2x^2+5x+2)/(x^2-4x+3)

$y = \frac{2 x^{2} + 5 x + 2}{x^{2} - 4 x + 3}$

Step 1

Find the x-intercepts.

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

$0 = \frac{2 x^{2} + 5 x + 2}{x^{2} - 4 x + 3}$

Solve the equation.

Set the numerator equal to zero.

$2 x^{2} + 5 x + 2 = 0$

Solve the equation for $x$ .

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) + 2 = 0$

Rewrite $5$ as $1$ plus $4$

$2 x^{2} + (1 + 4) x + 2 = 0$

Apply the distributive property.

$2 x^{2} + 1 x + 4 x + 2 = 0$

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} + x) + 4 x + 2 = 0$

Factor out the greatest common factor (GCF) from each group.

$x (2 x + 1) + 2 (2 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$(2 x + 1) (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x + 1 = 0$

$x + 2 = 0$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(2 x + 1) (x + 2) = 0$ true.

$x = - \frac{1}{2}, - 2$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{1}{2}, 0), (- 2, 0)$

Step 2

Find the y-intercepts.

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

$y = \frac{2 {(0)}^{2} + 5 (0) + 2}{{(0)}^{2} - 4 \cdot 0 + 3}$

Solve the equation.

Remove parentheses.

$y = \frac{2 \cdot 0^{2} + 5 (0) + 2}{{(0)}^{2} - 4 \cdot 0 + 3}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} + 5 (0) + 2}{0^{2} - 4 \cdot 0 + 3}$

Remove parentheses.

$y = \frac{2 {(0)}^{2} + 5 (0) + 2}{{(0)}^{2} - 4 \cdot 0 + 3}$

Simplify $\frac{2 {(0)}^{2} + 5 (0) + 2}{{(0)}^{2} - 4 \cdot 0 + 3}$ .

Simplify the numerator.

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

$y = \frac{2 \cdot 0 + 5 (0) + 2}{0^{2} - 4 \cdot 0 + 3}$

Multiply $2$ by $0$ .

$y = \frac{0 + 5 (0) + 2}{0^{2} - 4 \cdot 0 + 3}$

Multiply $5$ by $0$ .

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

Add $0$ and $0$ .

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

Add $0$ and $2$ .

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

Simplify the denominator.

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

$y = \frac{2}{0 - 4 \cdot 0 + 3}$

Multiply $- 4$ by $0$ .

$y = \frac{2}{0 + 0 + 3}$

Add $0$ and $0$ .

$y = \frac{2}{0 + 3}$

Add $0$ and $3$ .

$y = \frac{2}{3}$

y-intercept(s) in point form.

y-intercept(s): $(0, \frac{2}{3})$

Step 3

List the intersections.

x-intercept(s): $(- \frac{1}{2}, 0), (- 2, 0)$

y-intercept(s): $(0, \frac{2}{3})$

Step 4

Find the x and y Intercepts y=(2x^2+5x+2)/(x^2-4x+3)

Related Questions

Support LCMGCF.com by disabling your adblocker.