Find the Vertex y=4x^2+2x

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

Step 1

Rewrite the equation in vertex form.

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

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

$a = 4$

$b = 2$

$c = 0$

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$e = 0 - \frac{2^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

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

$e = 0 - \frac{4}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{4}{16}$

Cancel the common factor of $4$ and $16$ .

Factor $4$ out of $4$ .

$e = 0 - \frac{4 (1)}{16}$

Cancel the common factors.

Factor $4$ out of $16$ .

$e = 0 - \frac{4 \cdot 1}{4 \cdot 4}$

Cancel the common factor.

$e = 0 - \frac{4 \cdot 1}{4 \cdot 4}$

Rewrite the expression.

$e = 0 - \frac{1}{4}$

Subtract $\frac{1}{4}$ from $0$ .

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

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x + \frac{1}{4})}^{2} - \frac{1}{4}$ .

$4 {(x + \frac{1}{4})}^{2} - \frac{1}{4}$

Set $y$ equal to the new right side.

$y = 4 {(x + \frac{1}{4})}^{2} - \frac{1}{4}$

Step 2

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

$a = 4$

$h = - \frac{1}{4}$

$k = - \frac{1}{4}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{1}{4}, - \frac{1}{4})$

Step 4

Find the Vertex y=4x^2+2x

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