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The vertex of the positive parabola turns at point (-2, -11)
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)
y=aX2+bX+c ---> y=X2-6X+5vertex formula is -b/2a( b=-6,a=1)vertex=-(-6)/2*1=6/2=3 y=32-6(3)+5---->y=-4so vertex is (3,4)
(3, -21)
y = 2x2 + 4x - 3 This equation describes a parabola. Because it's first term is positive, we know that it goes infinitely upward, and has a minimum that occurs at it's vertex. You can find it's vertex by taking it's derivative and solving for zero: y' = 4x + 4 0 = 4x + 4 0 = x + 1 x = -1 y = (-1)2 + 4(-1) - 3 y = 1 - 4 - 3 y = -6 So the vertex is at (-1, -6), which means that y ≥ -6