Precalculus Examples

$- x^{2} + 6 x + 4$

Step 1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = 6$

$c = 4$

Step 2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{6}{2 \cdot - 1}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Cancel the common factor of $6$ and $2$ .

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Step 3.2.1.1

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot - 1}$

Step 3.2.1.2

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

Step 3.2.2

Multiply $- 1$ by $3$ .

$d = - 3$

Step 4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 4 - \frac{6^{2}}{4 \cdot - 1}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $6$ to the power of $2$ .

$e = 4 - \frac{36}{4 \cdot - 1}$

Step 4.2.1.2

Multiply $4$ by $- 1$ .

$e = 4 - \frac{36}{- 4}$

Step 4.2.1.3

Divide $36$ by $- 4$ .

$e = 4 - - 9$

Step 4.2.1.4

Multiply $- 1$ by $- 9$ .

$e = 4 + 9$

Step 4.2.2

Add $4$ and $9$ .

$e = 13$

Step 5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 3)}^{2} + 13$ .

$- {(x - 3)}^{2} + 13$