Find the x and y Intercepts y=(2x^2-3x+1)/(2x^2+8x+6)

$y = \frac{2 x^{2} - 3 x + 1}{2 x^{2} + 8 x + 6}$

Step 1

Find the x-intercepts.

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

$0 = \frac{2 x^{2} - 3 x + 1}{2 x^{2} + 8 x + 6}$

Solve the equation.

Set the numerator equal to zero.

$2 x^{2} - 3 x + 1 = 0$

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$2 x^{2} - 3 x + 1 = 0$

Rewrite $- 3$ as $- 1$ plus $- 2$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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

$2 x - 1 = 0$

$x - 1 = 0$

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 $x$ .

Add $1$ to 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 $1$ .

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

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

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

$y = \frac{2 {(0)}^{2} - 3 \cdot 0 + 1}{2 {(0)}^{2} + 8 (0) + 6}$

Solve the equation.

Remove parentheses.

$y = \frac{2 \cdot 0^{2} - 3 \cdot 0 + 1}{2 {(0)}^{2} + 8 (0) + 6}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} - 3 \cdot 0 + 1}{2 \cdot 0^{2} + 8 (0) + 6}$

Remove parentheses.

$y = \frac{2 {(0)}^{2} - 3 \cdot 0 + 1}{2 {(0)}^{2} + 8 (0) + 6}$

Simplify $\frac{2 {(0)}^{2} - 3 \cdot 0 + 1}{2 {(0)}^{2} + 8 (0) + 6}$ .

Simplify the numerator.

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

$y = \frac{2 \cdot 0 - 3 \cdot 0 + 1}{2 \cdot 0^{2} + 8 (0) + 6}$

Multiply $2$ by $0$ .

$y = \frac{0 - 3 \cdot 0 + 1}{2 \cdot 0^{2} + 8 (0) + 6}$

Multiply $- 3$ by $0$ .

$y = \frac{0 + 0 + 1}{2 \cdot 0^{2} + 8 (0) + 6}$

Add $0$ and $0$ .

$y = \frac{0 + 1}{2 \cdot 0^{2} + 8 (0) + 6}$

Add $0$ and $1$ .

$y = \frac{1}{2 \cdot 0^{2} + 8 (0) + 6}$

Simplify the denominator.

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

$y = \frac{1}{2 \cdot 0 + 8 (0) + 6}$

Multiply $2$ by $0$ .

$y = \frac{1}{0 + 8 (0) + 6}$

Multiply $8$ by $0$ .

$y = \frac{1}{0 + 0 + 6}$

Add $0$ and $0$ .

$y = \frac{1}{0 + 6}$

Add $0$ and $6$ .

$y = \frac{1}{6}$

y-intercept(s) in point form.

y-intercept(s): $(0, \frac{1}{6})$

Step 3

List the intersections.

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

y-intercept(s): $(0, \frac{1}{6})$

Step 4

Find the x and y Intercepts y=(2x^2-3x+1)/(2x^2+8x+6)

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