The function y = 6x^2 + x - 1 is a quadratic function, and its graph is a parabola. Since the coefficient of x^2 is positive, the parabola opens upwards, and its minimum value is at the vertex.

The x-coordinate of the vertex of a parabola given by y = ax^2 + bx + c is -b/2a.

So, for this function, the x-coordinate of the vertex is -b/2a = -1/(2*6) = -1/12.

Substitute x = -1/12 into the function to find the minimum value of y:

y = 6*(-1/12)^2 + (-1/12) - 1 = 6*(1/144) - 1/12 - 1 = 1/24 - 1/12 - 1 = -1 - 1/24 = -25/24.

So, the minimum value of y = 6x^2 + x - 1 is -25/24.

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The function y = 6x^2 + x - 1 is a quadratic function, and its graph is a parabola. Since the coefficient of x^2 is positive, the parabola opens upwards, and its minimum value is at the vertex.

The x-coordinate of the vertex of a parabola given by y = ax^2 + bx + c is -b/2a.

So, for this function, the x-coordinate of the vertex is -b/2a = -1/(2*6) = -1/12.

Substitute x = -1/12 into the function to find the minimum value of y:

y = 6*(-1/12)^2 + (-1/12) - 1 = 6*(1/144) - 1/12 - 1 = 1/24 - 1/12 - 1 = -1 - 1/24 = -25/24.

So, the minimum value of y = 6x^2 + x - 1 is -25/24.

Knowee AI · Accepted Answer

The function y = 6x^2 + x - 1 is a quadratic function, and its graph is a parabola. Since the coefficient of x^2 is positive, the parabola opens upwards, and its minimum value is at the vertex.

The x-coordinate of the vertex of a parabola given by y = ax^2 + bx + c is -b/2a.

So, for this function, the x-coordinate of the vertex is -b/2a = -1/(2*6) = -1/12.

Substitute x = -1/12 into the function to find the minimum value of y:

y = 6*(-1/12)^2 + (-1/12) - 1 = 6*(1/144) - 1/12 - 1 = 1/24 - 1/12 - 1 = -1 - 1/24 = -25/24.

So, the minimum value of y = 6x^2 + x - 1 is -25/24.

The minimum value of y=6x 2 +x−1 is:

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