Find the x and y Intercepts (2x^2+10x-12)/(x^2+x-6)

$\frac{2 x^{2} + 10 x - 12}{x^{2} + x - 6}$

Step 1

Write $\frac{2 x^{2} + 10 x - 12}{x^{2} + x - 6}$ as an equation.

$y = \frac{2 x^{2} + 10 x - 12}{x^{2} + x - 6}$

Step 2

Find the x-intercepts.

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

$0 = \frac{2 x^{2} + 10 x - 12}{x^{2} + x - 6}$

Solve the equation.

Set the numerator equal to zero.

$2 x^{2} + 10 x - 12 = 0$

Solve the equation for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 10 x - 12$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 10 x - 12 = 0$

Factor $2$ out of $10 x$ .

$2 (x^{2}) + 2 (5 x) - 12 = 0$

Factor $2$ out of $- 12$ .

$2 x^{2} + 2 (5 x) + 2 \cdot - 6 = 0$

Factor $2$ out of $2 x^{2} + 2 (5 x)$ .

$2 (x^{2} + 5 x) + 2 \cdot - 6 = 0$

Factor $2$ out of $2 (x^{2} + 5 x) + 2 \cdot - 6$ .

$2 (x^{2} + 5 x - 6) = 0$

Factor.

Factor $x^{2} + 5 x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $5$ .

$- 1, 6$

Write the factored form using these integers.

$2 ((x - 1) (x + 6)) = 0$

Remove unnecessary parentheses.

$2 (x - 1) (x + 6) = 0$

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

$x - 1 = 0$

$x + 6 = 0$

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.

$x = 1$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $2 (x - 1) (x + 6) = 0$ true.

$x = 1, - 6$

x-intercept(s) in point form.

x-intercept(s): $(1, 0), (- 6, 0)$

Step 3

Find the y-intercepts.

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

$y = \frac{2 {(0)}^{2} + 10 (0) - 12}{{(0)}^{2} + 0 - 6}$

Solve the equation.

Remove parentheses.

$y = \frac{2 {(0)}^{2} + 10 (0) - 12}{{(0)}^{2} + 0 - 6}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} + 10 (0) - 12}{{(0)}^{2} + 0 - 6}$

Remove parentheses.

$y = \frac{2 \cdot 0^{2} + 10 (0) - 12}{0^{2} + 0 - 6}$

Remove parentheses.

$y = \frac{2 {(0)}^{2} + 10 (0) - 12}{{(0)}^{2} + 0 - 6}$

Simplify $\frac{2 {(0)}^{2} + 10 (0) - 12}{{(0)}^{2} + 0 - 6}$ .

Simplify the numerator.

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

$y = \frac{2 \cdot 0 + 10 (0) - 12}{0^{2} + 0 - 6}$

Multiply $2$ by $0$ .

$y = \frac{0 + 10 (0) - 12}{0^{2} + 0 - 6}$

Multiply $10$ by $0$ .

$y = \frac{0 + 0 - 12}{0^{2} + 0 - 6}$

Add $0$ and $0$ .

$y = \frac{0 - 12}{0^{2} + 0 - 6}$

Subtract $12$ from $0$ .

$y = \frac{- 12}{0^{2} + 0 - 6}$

Simplify the denominator.

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

$y = \frac{- 12}{0 + 0 - 6}$

Add $0$ and $0$ .

$y = \frac{- 12}{0 - 6}$

Subtract $6$ from $0$ .

$y = \frac{- 12}{- 6}$

Divide $- 12$ by $- 6$ .

$y = 2$

y-intercept(s) in point form.

y-intercept(s): $(0, 2)$

Step 4

List the intersections.

x-intercept(s): $(1, 0), (- 6, 0)$

y-intercept(s): $(0, 2)$

Step 5

Find the x and y Intercepts (2x^2+10x-12)/(x^2+x-6)

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