Find the Vertex y+2x^2-4x-6=0

$y + 2 x^{2} - 4 x - 6 = 0$

Step 1

Rewrite the equation in vertex form.

Move all terms not containing $y$ to the right side of the equation.

Subtract $2 x^{2}$ from both sides of the equation.

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

Add $4 x$ to both sides of the equation.

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

Add $6$ to both sides of the equation.

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

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

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

$a = - 2$

$b = 4$

$c = 6$

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

$d = \frac{4}{2 \cdot - 2}$

Simplify the right side.

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $2$ out of $2$ .

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

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

$d = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$d = - 1$

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

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

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

Cancel the common factors.

Factor $4$ out of $4 \cdot - 2$ .

$e = 6 - \frac{4 \cdot 4}{4 (- 2)}$

Cancel the common factor.

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

Rewrite the expression.

$e = 6 - \frac{4}{- 2}$

Cancel the common factor of $4$ and $- 2$ .

Factor $2$ out of $4$ .

$e = 6 - \frac{2 (2)}{- 2}$

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$e = 6 - (- 1 \cdot 2)$

Multiply $- (- 1 \cdot 2)$ .

Multiply $- 1$ by $2$ .

$e = 6 - - 2$

Multiply $- 1$ by $- 2$ .

$e = 6 + 2$

Add $6$ and $2$ .

$e = 8$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 2 {(x - 1)}^{2} + 8$ .

$- 2 {(x - 1)}^{2} + 8$

Set $y$ equal to the new right side.

$y = - 2 {(x - 1)}^{2} + 8$

Step 2

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

$a = - 2$

$h = 1$

$k = 8$

Step 3

Find the vertex $(h, k)$ .

$(1, 8)$

Step 4

Find the Vertex y+2x^2-4x-6=0

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