Find the Vertex 4x^2+16x+4

$4 x^{2} + 16 x + 4$

Step 1

Rewrite the equation in vertex form.

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

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

$a = 4$

$b = 16$

$c = 4$

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

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $4$ .

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

Divide $256$ by $16$ .

$e = 4 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 4 - 16$

Subtract $16$ from $4$ .

$e = - 12$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x + 2)}^{2} - 12$ .

$4 {(x + 2)}^{2} - 12$

Set $y$ equal to the new right side.

$y = 4 {(x + 2)}^{2} - 12$

Step 2

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

$a = 4$

$h = - 2$

$k = - 12$

Step 3

Find the vertex $(h, k)$ .

$(- 2, - 12)$

Step 4

Find the Vertex 4x^2+16x+4

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