Find the Vertex Form y=8x^2-16x+27

$y = 8 x^{2} - 16 x + 27$

Step 1

Complete the square for $8 x^{2} - 16 x + 27$ .

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

$a = 8$

$b = - 16$

$c = 27$

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

$d = \frac{- 16}{2 \cdot 8}$

Simplify the right side.

Cancel the common factor of $- 16$ and $2$ .

Factor $2$ out of $- 16$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 8$ .

$d = \frac{2 \cdot - 8}{2 (8)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{- 8}{8}$

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

Factor $8$ out of $- 8$ .

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

Cancel the common factors.

Factor $8$ out of $8$ .

$d = \frac{8 \cdot - 1}{8 (1)}$

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ 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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 27 - \frac{{(- 16)}^{2}}{4 \cdot 8}$

Simplify the right side.

Simplify each term.

Raise $- 16$ to the power of $2$ .

$e = 27 - \frac{256}{4 \cdot 8}$

Multiply $4$ by $8$ .

$e = 27 - \frac{256}{32}$

Divide $256$ by $32$ .

$e = 27 - 1 \cdot 8$

Multiply $- 1$ by $8$ .

$e = 27 - 8$

Subtract $8$ from $27$ .

$e = 19$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $8 {(x - 1)}^{2} + 19$ .

$8 {(x - 1)}^{2} + 19$

Step 2

Set $y$ equal to the new right side.

$y = 8 {(x - 1)}^{2} + 19$

Find the Vertex Form y=8x^2-16x+27

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