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Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
x = +/- sqrt(y/3)
No. Y=3X2+1 is a Binomial Equation.
Yes.
(3x + 3y)(x + y) or 3(x + y)2
3x2-y=6 3x2-y=6
Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
3 * x * x * y
x = +/- sqrt(y/3)
72
You can find the x-coordinate of it's vertex by taking it's derivative and solving for zero: y = -3x2 + 12x - 5 y' = -6x + 12 0 = -6x + 12 6x = 12 x = 2 Now that we have it's x coordinate, we can plug it back into the original equation to find it's y coordinate: y = -3x2 + 12x - 5 y = -3(2)2 + 12(2) + 5 y = -12 + 24 + 5 y = 17 So the vertex of the parabola y = -3x2 + 12x - 5 occurs at the point (2, 17).
No. Y=3X2+1 is a Binomial Equation.
Yes.
(3x + 3y)(x + y) or 3(x + y)2
The figures are exactly the same, but every point on the first graph is exactly 13 below the corresponding point on the second one.
(x2 - 2)(y - 3)
y=2x2-3x2-12x+5=0