# What is the vertex form of y=7x^2 +3x + 5 ?

$y = 7 {\left(x + \frac{3}{14}\right)}^{2} + \frac{917}{196}$
The vertex form of a quadratic equation $y = a {x}^{2} + b x + c$ is $y = a {\left(x + m\right)}^{2} + n$, where $m = \frac{b}{2 a}$ and $n = - a {\left(\frac{b}{2 a}\right)}^{2} + c$
Then the vertex is at the point where the bracketed expression is zero and is therefore $\left(- m , n\right)$
Therefore $y = 7 {\left(x + \frac{3}{14}\right)}^{2} - 7 \cdot \frac{9}{196} + 5$
$y = 7 {\left(x + \frac{3}{14}\right)}^{2} - \frac{63 + 980}{196}$
$y = 7 {\left(x + \frac{3}{14}\right)}^{2} + \frac{917}{196}$