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mathhelper y = 2x^2 + 8x + 9
several methods:
easiest way, the x of the vertex of y = ax^2 + bx + c is -b/(2a)
= -8/4 = -2
then y = 2(-2)^2 + 8(-2) + 9 = 1
vertex is (-2,1)
by completing the square:
y = 2(x^2 + 4x + ... - ....) + 9
= 2(x^2 + 4x + 4 - 4) + 9
= 2( (x+2)^2 - 4) + 9
= 2(x+2)^2 - 8 + 9
= 2(x+2)^2 + 1 <===== vertex form , so vertex is (-2,1)
not sure if you know calculus, but here it is....
dy/dx = 4x + 8 = 0 at the vertex
4x + 8 = 0
x = -2, then y = 1 just like before