Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
if you rearrange it, it becomes 4y + 8 = 3x2 -2x +1 4y = 3x2 -2x -7 y= (3x2 -2x -7)/4 which is a parabola
x = +/- sqrt(y/3)
No. Y=3X2+1 is a Binomial Equation.
Yes.
3x2-y=6 3x2-y=6
Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
3x2-27 is equal to -21.
if you rearrange it, it becomes 4y + 8 = 3x2 -2x +1 4y = 3x2 -2x -7 y= (3x2 -2x -7)/4 which is a parabola
3 * x * x * y
3x2-27 is equal to -21.
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.
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
Yes.