Answer
$y=4\left(x+\frac{7}{8}\right)^2-\frac{49}{16}$
Work Step by Step
Recall:
(1) The vertex form of $y=ax^2+bx+c$ is $y=a(x-h)^2+k$ where $(h, k)$ is the vertex.
(2) The vertex of the graph of $y=ax^2+bx+c$ is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.
The given function has $a=4$ and $b=7$.
Thus, the $x$-coordinate of the vertex is:
\begin{align*}
x&=-\frac{b}{2a}\\\\
x&=-\frac{7}{2(4)}\\\\
x&=-\frac{7}{8}\\\\
\end{align*}
Find the $y$-coordinate of the vertex by substituting $-\frac{7}{8}$ to $x$ to obtain:
\begin{align*}
y&=4x^2+7x\\\\
y&=4\left(-\frac{7}{8}\right)^2+7\left(-\frac{7}{8}\right)\\\\
y&=4\left(\frac{49}{64}\right)-\left(\frac{49}{8}\right)\\\\
y&=\frac{49}{16}-\frac{49}{8}\\\\
y&=\frac{49}{16}-\frac{98}{16}\\\\
y&=-\frac{49}{16}
\end{align*}
The vertex is at $(-1, 4)$ so $h=-1$ and $k=4$.
Substitute the values of $a, h,$ and $k$ into the vertex form in the recall part (1) above to obtain:
$$y=4\left(x+\frac{7}{8}\right)^2-\frac{49}{16}$$
