# What is the vertex form of f(x) = x^2+4x+6?

Feb 13, 2016

$y = {\left(x + 2\right)}^{2} + 2$

#### Explanation:

the standard form of a quadratic function is $y = a {x}^{2} + b x + c$

here $f \left(x\right) = {x}^{2} + 4 x + 6$

and by comparison : a = 1 , b = 4 and c = 6

in vertex form the equation is: $y = a {\left(x - h\right)}^{2} + k$

where ( h , k ) are the coords of the vertex.

the x-coord of vertex $= - \frac{b}{2 a} = - \frac{4}{2} = - 2$

and y-coord. =${\left(- 2\right)}^{2} + 4 \left(- 2\right) + 6 = 4 - 8 + 6 = 2$

now (h , k) =(-2 , 2) and a = 1

$\Rightarrow y = {\left(x + 2\right)}^{2} + 2$