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What is the x-coordinate of the of the vertex of y equals -3x2 plus 12x-5?
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).
How do you find the x- coordinates of the turning points of y equals x3-2x2-5x plus 6?
Differentiate the function with respect to x: d/dx (x3 - 2x2 - 5x + 6) = 3x2 - 4x - 5 Set this derivative = 0 and solve. 3x2 - 4x - 5 = 0 implies that x = -0.7863 or 2.1196 (to 4 dp)
Is 3x - 2 equals -3x2 a quadratic equation?
Yes. (Assuming that -3x2 is the best representation of 3x2 that this browser will allow.)
What does 6-y plus 5y-6y?
6 - y + 5y - 6y = 0 6 = y -5y + 6y 6 = -4y + 6y 6 = 2y 6/2 = y 3 = y
How do you graph y -6?
y = -6 is a horizontal line that cuts the y-axis at -6.
