Find the Vertex y=6-x^2

$y = 6 - x^{2}$

Rewrite the equation in vertex form.

Reorder $6$ and $- x^{2}$ .

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

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

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

$a = - 1$

$b = 0$

$c = 6$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{0}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{0}{- 1}$ .

$d = - 1 \cdot 0$

Rewrite $- 1 \cdot 0$ as $- 0$ .

$d = - 0$

Multiply $- 1$ by $0$ .

$d = 0$

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

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

$e = 6 - \frac{0^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

$e = 6 - \frac{0}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 6 - \frac{0}{- 4}$

Divide $0$ by $- 4$ .

$e = 6 - 0$

Multiply $- 1$ by $0$ .

$e = 6 + 0$

Add $6$ and $0$ .

$e = 6$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 0)}^{2} + 6$ .

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

Set $y$ equal to the new right side.

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

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

$a = - 1$

$h = 0$

$k = 6$

Find the vertex $(h, k)$ .

$(0, 6)$

Find the Vertex y=6-x^2

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