Let f(x)=x2+4x+12 . what is the vertex form of f(x)? what is the minimum value of f(x)?

The answer is  y=(x−2)2−16 but the explanation is  the standard form of a quadratic function is y=ax2+bx+cthe equation here  y=x2−4x−12 is of this form by comparison : a = 1 , b = -4 and c = -12the vertex form of the quadratic function isy=a(x−h)2+k  where (h , k ) are the coords of the vertex.the x-coord of the vertex =  −b2a=−−42=2  substitute x = 2 into original function for y-coord.y =  (2)2−4(2)−12=4−8−12=−16  hence (h , k )= (2 , -16 ) and a = 1

⇒y=(x−2)2−16 

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