Question

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

Answer 1

`y = (x+2)^2 + 2`

Explanation:

the standard form of a quadratic function is `y = ax^2 + bx + c`

here `f(x) = x^2 + 4x + 6`

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

in vertex form the equation is: `y = a(x-h)^2 + k`

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

the x-coord of vertex `= -b/(2a) = -4/2 = - 2`

and y-coord. =`(-2)^2 + 4(-2) +6 = 4 - 8 + 6 = 2`

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

`rArr y = (x+2)^2 + 2`