Find the Vertex -x^2-6x+4

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Algebra

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

Set the polynomial equal to $y$ to find the properties of the parabola.

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

Rewrite the equation in vertex form.

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Complete the square for $- x^{2} - 6 x + 4$ .

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

$a = - 1, b = - 6, c = 4$

Consider the vertex form of a parabola.

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

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

$d = - \frac{6}{2 (- 1)}$

Simplify the right side.

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

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Factor $2$ out of $6$ .

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

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

$d = - (- 1 \cdot 3)$

Multiply.

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Multiply $- 1$ by $3$ .

$d = 3$

Multiply $- 1$ by $- 3$ .

$d = 3$

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

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Simplify each term.

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

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

Multiply $4$ by $- 1$ .

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

Divide $36$ by $- 4$ .

$e = 4 + 9$

Multiply $- 1$ by $- 9$ .

$e = 4 + 9$

Add $4$ and $9$ .

$e = 13$

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

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

Set $y$ equal to the new right side.

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

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = - 1$

$h = - 3$

$k = 13$

Find the vertex $(h, k)$ .

$(- 3, 13)$

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