Precalculus Examples

$y = \frac{1}{4} x^{2} - 2 x - 12$

Step 1

Set $\frac{1}{4} x^{2} - 2 x - 12$ equal to $0$ .

$\frac{1}{4} x^{2} - 2 x - 12 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Combine $\frac{1}{4}$ and $x^{2}$ .

$\frac{x^{2}}{4} - 2 x - 12 = 0$

Step 2.2

Multiply each term in $\frac{x^{2}}{4} - 2 x - 12 = 0$ by $4$ to eliminate the fractions.

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Step 2.2.1

Multiply each term in $\frac{x^{2}}{4} - 2 x - 12 = 0$ by $4$ .

$\frac{x^{2}}{4} \cdot 4 - 2 x \cdot 4 - 12 \cdot 4 = 0 \cdot 4$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Cancel the common factor of $4$ .

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Step 2.2.2.1.1.1

Cancel the common factor.

$\frac{x^{2}}{4} \cdot 4 - 2 x \cdot 4 - 12 \cdot 4 = 0 \cdot 4$

Step 2.2.2.1.1.2

Rewrite the expression.

$x^{2} - 2 x \cdot 4 - 12 \cdot 4 = 0 \cdot 4$

Step 2.2.2.1.2

Multiply $4$ by $- 2$ .

$x^{2} - 8 x - 12 \cdot 4 = 0 \cdot 4$

Step 2.2.2.1.3

Multiply $- 12$ by $4$ .

$x^{2} - 8 x - 48 = 0 \cdot 4$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Multiply $0$ by $4$ .

$x^{2} - 8 x - 48 = 0$

Step 2.3

Factor $x^{2} - 8 x - 48$ using the AC method.

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Step 2.3.1

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 $- 48$ and whose sum is $- 8$ .

$- 12, 4$

Step 2.3.2

Write the factored form using these integers.

$(x - 12) (x + 4) = 0$

Step 2.4

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

$x - 12 = 0$

$x + 4 = 0$

Step 2.5

Set $x - 12$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

Step 2.5.2

Add $12$ to both sides of the equation.

$x = 12$

Step 2.6

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.6.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.7

The final solution is all the values that make $(x - 12) (x + 4) = 0$ true.

$x = 12, - 4$

Step 3