Find the Vertex Form y=2x^2+8x+1

$y = 2 x^{2} + 8 x + 1$

Step 1

Complete the square for $2 x^{2} + 8 x + 1$ .

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

$a = 2$

$b = 8$

$c = 1$

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

$d = \frac{8}{2 \cdot 2}$

Simplify the right side.

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

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

Cancel the common factors.

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

$d = \frac{2 \cdot 4}{2 (2)}$

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $2$ out of $4$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{2}{1}$

Divide $2$ by $1$ .

$d = 2$

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

$e = 1 - \frac{8^{2}}{4 \cdot 2}$

Simplify the right side.

Simplify each term.

Raise $8$ to the power of $2$ .

$e = 1 - \frac{64}{4 \cdot 2}$

Multiply $4$ by $2$ .

$e = 1 - \frac{64}{8}$

Divide $64$ by $8$ .

$e = 1 - 1 \cdot 8$

Multiply $- 1$ by $8$ .

$e = 1 - 8$

Subtract $8$ from $1$ .

$e = - 7$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $2 {(x + 2)}^{2} - 7$ .

$2 {(x + 2)}^{2} - 7$

Step 2

Set $y$ equal to the new right side.

$y = 2 {(x + 2)}^{2} - 7$

Find the Vertex Form y=2x^2+8x+1

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