Find the Vertex y=-2x^2-4x-1

$y = - 2 x^{2} - 4 x - 1$

Rewrite the equation in vertex form.

Complete the square for $- 2 x^{2} - 4 x - 1$ .

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

$a = - 2$

$b = - 4$

$c = - 1$

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 $- 2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

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

Simplify the right side.

Simplify each term.

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

Rewrite $- 4$ as $- 1 (4)$ .

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

Apply the product rule to $- 1 (4)$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply $4^{2}$ by $1$ .

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

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

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

Cancel the common factors.

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

$e = - 1 - \frac{4 \cdot 4}{4 (- 2)}$

Cancel the common factor.

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

Rewrite the expression.

$e = - 1 - \frac{4}{- 2}$

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

Factor $2$ out of $4$ .

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

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

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

Multiply $- 1$ by $2$ .

$e = - 1 - - 2$

Multiply $- 1$ by $- 2$ .

$e = - 1 + 2$

Add $- 1$ and $2$ .

$e = 1$

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

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

Set $y$ equal to the new right side.

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

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

$a = - 2$

$h = - 1$

$k = 1$

Find the vertex $(h, k)$ .

$(- 1, 1)$

Find the Vertex y=-2x^2-4x-1

Related Questions

Support LCMGCF.com by disabling your adblocker.