# How do you write the quadratic function f(x) = x^2 -4x + 7 in vertex form?

$f \left(x\right) = {\left(x - 2\right)}^{2} + 3$
If a standard quadratic has the form $y = a {x}^{2} + b x + c$, it can be rewritten into vertex form by completing the square. This means that we use $\frac{b}{2} a$ to get an expression a(x+b/2a)^2 which equals $a \left({x}^{2} + \frac{b}{a} \cdot x + {b}^{2} / \left(4 {a}^{2}\right)\right)$. We then subtract ${b}^{2} / \left(4 {a}^{2}\right)$ and add$c$ to finish the equation.
$f \left(x\right) = {\left(x - 2\right)}^{2} - 4 + 7$
$f \left(x\right) = {\left(x - 2\right)}^{2} + 3$
This is vertex form, from which you can get that the vertex is at $\left(2 , 3\right)$ - th epoint at which teh bracketed term is zero and therefore the expression is at its minimum value.