Find the x and y Intercepts f(x)=(2x-2)/(2x^2+3x-9)

$f (x) = \frac{2 x - 2}{2 x^{2} + 3 x - 9}$

Step 1

Find the x-intercepts.

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

$0 = \frac{2 x - 2}{2 x^{2} + 3 x - 9}$

Solve the equation.

Set the numerator equal to zero.

$2 x - 2 = 0$

Solve the equation for $x$ .

Add $2$ to both sides of the equation.

$2 x = 2$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $2$ by $2$ .

$x = 1$

x-intercept(s) in point form.

x-intercept(s): $(1, 0)$

Step 2

Find the y-intercepts.

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

$y = \frac{2 (0) - 2}{2 {(0)}^{2} + 3 (0) - 9}$

Solve the equation.

Remove parentheses.

$y = \frac{2 (0) - 2}{2 \cdot 0^{2} + 3 (0) - 9}$

Remove parentheses.

$y = \frac{2 (0) - 2}{2 {(0)}^{2} + 3 (0) - 9}$

Simplify $\frac{2 (0) - 2}{2 {(0)}^{2} + 3 (0) - 9}$ .

Simplify the numerator.

Multiply $2$ by $0$ .

$y = \frac{0 - 2}{2 \cdot 0^{2} + 3 (0) - 9}$

Subtract $2$ from $0$ .

$y = \frac{- 2}{2 \cdot 0^{2} + 3 (0) - 9}$

Simplify the denominator.

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

$y = \frac{- 2}{2 \cdot 0 + 3 (0) - 9}$

Multiply $2$ by $0$ .

$y = \frac{- 2}{0 + 3 (0) - 9}$

Multiply $3$ by $0$ .

$y = \frac{- 2}{0 + 0 - 9}$

Add $0$ and $0$ .

$y = \frac{- 2}{0 - 9}$

Subtract $9$ from $0$ .

$y = \frac{- 2}{- 9}$

Dividing two negative values results in a positive value.

$y = \frac{2}{9}$

y-intercept(s) in point form.

y-intercept(s): $(0, \frac{2}{9})$

Step 3

List the intersections.

x-intercept(s): $(1, 0)$

y-intercept(s): $(0, \frac{2}{9})$

Step 4

Find the x and y Intercepts f(x)=(2x-2)/(2x^2+3x-9)

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