If your function is 4x^2+16x+16=0 then x=-2, if it is 4x^2-16x+16=0, then x=2
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It is; (2x+4)^2.
True
x3 + 4x2 + 16x + 64 =x2*(x + 4) + 16(x + 4) = (x + 4)*(x2 + 16) which has no further real factors.
5x - 16 = 3x2x - 16 = 02x = 16x = 8
f(x) = -4x2 - 16x - 11 a = -4, b = -16, c = -11 x-coordinate of the vertex = -b/2a = -(-16)/2(-4) = 16/-8 = -2 y-coordinate = f(-2) = -4(-2)2 -16(-2) - 11 = -16 + 32 - 11 = 5 vertex is (-2, 5)