Find the Vertex Form f(x)=4x^2-16x+16

$f (x) = 4 x^{2} - 16 x + 16$

Step 1

Write $f (x) = 4 x^{2} - 16 x + 16$ as an equation.

$y = 4 x^{2} - 16 x + 16$

Step 2

Complete the square for $4 x^{2} - 16 x + 16$ .

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

$a = 4$

$b = - 16$

$c = 16$

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

$d = \frac{- 16}{2 \cdot 4}$

Simplify the right side.

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

Factor $2$ out of $- 16$ .

$d = \frac{2 \cdot - 8}{2 \cdot 4}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 8}{2 (4)}$

Cancel the common factor.

$d = \frac{2 \cdot - 8}{2 \cdot 4}$

Rewrite the expression.

$d = \frac{- 8}{4}$

Cancel the common factor of $- 8$ and $4$ .

Factor $4$ out of $- 8$ .

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

Cancel the common factors.

Factor $4$ out of $4$ .

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

Cancel the common factor.

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

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

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

Simplify the right side.

Simplify each term.

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

$e = 16 - \frac{256}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 16 - \frac{256}{16}$

Divide $256$ by $16$ .

$e = 16 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 16 - 16$

Subtract $16$ from $16$ .

$e = 0$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x - 2)}^{2} + 0$ .

$4 {(x - 2)}^{2} + 0$

Step 3

Set $y$ equal to the new right side.

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

Find the Vertex Form f(x)=4x^2-16x+16

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