What is the vertex of the parbola y equals x2-4x plus 12?

Answer 1

Differentiate and equate to zero

Hence

y = x^2 + 4x - 12

dy/dx = 2x + 4 = 0

2x = -4

x = -2

Hence

y = (-2)^2 +4(-2) - 12

y = 4 - 8 - 12

y = - 16

Hence the vertex is (x,y,) = ( -2,-16)

Answer 2

Y = X2 - 4X + 12

set to 0

X2 - 4X + 12 = 0

subtract 12 from each side

X2 - 4X = - 12

now, take the linear term ( - 4 ), halve it, square it and add it to both sides

X2 - 4X + 4 = - 12 + 4

factor the left side, gather terms on the right

(X - 2)2 = - 8

add 8 to each side

(X - 2)2 + 8 = 0

------------------------

(2, 8)

------------the vertex of this function