Find the Vertex y=-2x^2+5

$y = - 2 x^{2} + 5$

Step 1

Rewrite the equation in vertex form.

Complete the square for $- 2 x^{2} + 5$ .

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

$a = - 2$

$b = 0$

$c = 5$

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 $d = \frac{b}{2 a}$ .

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

Simplify the right side.

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

Factor $2$ out of $0$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot - 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $2$ out of $0$ .

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

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 5 - \frac{0^{2}}{4 \cdot - 2}$

Simplify the right side.

Simplify each term.

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

$e = 5 - \frac{0}{4 \cdot - 2}$

Multiply $4$ by $- 2$ .

$e = 5 - \frac{0}{- 8}$

Divide $0$ by $- 8$ .

$e = 5 - 0$

Multiply $- 1$ by $0$ .

$e = 5 + 0$

Add $5$ and $0$ .

$e = 5$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 2 {(x + 0)}^{2} + 5$ .

$- 2 {(x + 0)}^{2} + 5$

Set $y$ equal to the new right side.

$y = - 2 {(x + 0)}^{2} + 5$

Step 2

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

$a = - 2$

$h = 0$

$k = 5$

Step 3

Find the vertex $(h, k)$ .

$(0, 5)$

Step 4

Find the Vertex y=-2x^2+5

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