Find the Vertex y=4x^2+7x

$y = 4 x^{2} + 7 x$

Rewrite the equation in vertex form.

Complete the square for $4 x^{2} + 7 x$ .

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

$a = 4$

$b = 7$

$c = 0$

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

$d = \frac{7}{2 \cdot 4}$

Multiply $2$ by $4$ .

$d = \frac{7}{8}$

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

$e = 0 - \frac{7^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

Raise $7$ to the power of $2$ .

$e = 0 - \frac{49}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{49}{16}$

Subtract $\frac{49}{16}$ from $0$ .

$e = - \frac{49}{16}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x + \frac{7}{8})}^{2} - \frac{49}{16}$ .

$4 {(x + \frac{7}{8})}^{2} - \frac{49}{16}$

Set $y$ equal to the new right side.

$y = 4 {(x + \frac{7}{8})}^{2} - \frac{49}{16}$

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

$a = 4$

$h = - \frac{7}{8}$

$k = - \frac{49}{16}$

Find the vertex $(h, k)$ .

$(- \frac{7}{8}, - \frac{49}{16})$