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

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Algebra

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

Isolate $y$ to the left side of the equation.

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Rewrite the equation as $y - 1 = x^{2} + 8 x$ .

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

Add $1$ to both sides of the equation.

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

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

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

$a = 1$

$b = 8$

$c = 1$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{8}{2 \cdot 1}$

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

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Factor $2$ out of $8$ .

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Cancel the common factors.

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Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

$d = \frac{2 \cdot 4}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{4}{1}$

Divide $4$ by $1$ .

$d = 4$

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

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

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

Simplify the right side.

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Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Divide $64$ by $4$ .

$e = 1 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 1 - 16$

Subtract $16$ from $1$ .

$e = - 15$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 4)}^{2} - 15$ .

${(x + 4)}^{2} - 15$

Set $y$ equal to the new right side.

$y = {(x + 4)}^{2} - 15$

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