What is the vertex form of f(x) = x^2+4x+6f(x)=x2+4x+6?
1 Answer
Feb 13, 2016
y = (x+2)^2 + 2y=(x+2)2+2
Explanation:
the standard form of a quadratic function is
y = ax^2 + bx + cy=ax2+bx+c here
f(x) = x^2 + 4x + 6f(x)=x2+4x+6 and by comparison : a = 1 , b = 4 and c = 6
in vertex form the equation is:
y = a(x-h)^2 + k 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 =−b2a=−42=−2 and y-coord. =
(-2)^2 + 4(-2) +6 = 4 - 8 + 6 = 2(−2)2+4(−2)+6=4−8+6=2 now (h , k) =(-2 , 2) and a = 1
rArr y = (x+2)^2 + 2 ⇒y=(x+2)2+2
