Precalculus Examples

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

Step 1

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

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

Step 2

Find the x-intercepts.

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Step 2.1

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

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

Step 2.2

Solve the equation.

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Step 2.2.1

Set the numerator equal to zero.

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

Step 2.2.2

Solve the equation for $x$ .

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Step 2.2.2.1

Factor the left side of the equation.

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Step 2.2.2.1.1

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

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Step 2.2.2.1.1.1

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

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

Step 2.2.2.1.1.2

Factor $2$ out of $2 x$ .

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

Step 2.2.2.1.1.3

Factor $2$ out of $- 12$ .

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

Step 2.2.2.1.1.4

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

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

Step 2.2.2.1.1.5

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

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

Step 2.2.2.1.2

Factor.

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Step 2.2.2.1.2.1

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

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Step 2.2.2.1.2.1.1

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 $1$ .

$- 2, 3$

Step 2.2.2.1.2.1.2

Write the factored form using these integers.

$2 ((x - 2) (x + 3)) = 0$

Step 2.2.2.1.2.2

Remove unnecessary parentheses.

$2 (x - 2) (x + 3) = 0$

Step 2.2.2.2

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

$x - 2 = 0$

$x + 3 = 0$

Step 2.2.2.3

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

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Step 2.2.2.3.1

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

$x - 2 = 0$

Step 2.2.2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.2.2.4

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.2.2.4.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.2.2.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.2.2.5

The final solution is all the values that make $2 (x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Step 2.3

x-intercept(s) in point form.

x-intercept(s): $(2, 0), (- 3, 0)$

Step 3

Find the y-intercepts.

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Step 3.1

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

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

Step 3.2

Solve the equation.

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Step 3.2.1

Remove parentheses.

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

Step 3.2.2

Remove parentheses.

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

Step 3.2.3

Remove parentheses.

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

Step 3.2.4

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

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Step 3.2.4.1

Simplify the numerator.

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Step 3.2.4.1.1

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

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

Step 3.2.4.1.2

Multiply $2$ by $0$ .

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

Step 3.2.4.1.3

Multiply $2$ by $0$ .

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

Step 3.2.4.1.4

Add $0$ and $0$ .

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

Step 3.2.4.1.5

Subtract $12$ from $0$ .

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

Step 3.2.4.2

Simplify the denominator.

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Step 3.2.4.2.1

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

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

Step 3.2.4.2.2

Multiply $- 1$ by $0$ .

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

Step 3.2.4.2.3

Add $0$ and $0$ .

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

Step 3.2.4.2.4

Subtract $6$ from $0$ .

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

Step 3.2.4.3

Divide $- 12$ by $- 6$ .

$y = 2$

Step 3.3

y-intercept(s) in point form.

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

Step 4

List the intersections.

x-intercept(s): $(2, 0), (- 3, 0)$

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

Step 5