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Q: What is the vertex of the parabola for the equation 3x2 6x 5?

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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).

y2-3x2+6x+6y= 18 is in standard form. The vertex form would be (y+3)2/24 - (x-1)2/8 = 1

The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.

Vertex = (3, - 2)Put in vertex form.(X - 3)2 + 2X2 - 6X + 9 + 2 = 0X2 - 6X + 11 = 0=============The coefficeint of the squared term is 1. My TI-84 confirms the (4, 3) intercept of the parabola and the 11 Y intercept shown by the function.

The value of x is 2

From what i understand the equation is like this; (3 * 2 ) + 6x + 24 = 0 6 + 6x +24 = 0 6x = - 24 - 6 6x = - 30 x = - 5

Parabola: 1+12x-6x^2 Factorizing: -6(x^2 -2x -1/6) Completing the square: -6((x-1)^2 -1 -1/6) => -6(x-1)^2 +7 Vertex of parabola is at: (1, 7)

20 and the vertex of the parabola is at (3, 20)

The question does not contain an equation: only an expression. An expression cannot have a vertex form.

In this case, the discriminant is less than zero and the graph of this parabola lies above the x-axis. It never crosses.

-22

The derivative if a function is basically it's slope, or its rate of change. An example is the function y = 4x - 6. This is a line with a slope of 4. The derivative is y' = 4. Another example is the function y = 3x2. This is a parabola with a vertex at (0,0). Its derivative is y' = 6x. At x = 0, the slope of the parabola is 6*0, which is 0, since this is the vertex of the parabola. To the left, at x is -4 for example, the derivative (and therefore slope) is negative. To the right, at x = 5 for example, the derivative is positive. The farther away from the vertex, the greater the value of the derivative so the the slope of the function increases as you move away from the vertex (it gets steeper).

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