Algebraically analyze the graph of the function f(x)= -6x^2+5x+2 . Find the vertex, axis of symmetry, range, minimum or maximum value, increasing and decreasing intervals, and zeros. Write the Quadratic equation in standard form. [Standard form is: f(x)=a(x-h)^2+k ]

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To algebraically analyze the graph of the function f(x) = -6x^2 + 5x + 2, we need to find the vertex, axis of symmetry, range, minimum or maximum value, increasing and decreasing intervals, and zeros. Additionally, we need to write the quadratic equation in standard form.

1. Vertex: The vertex of a quadratic function is given by the formula (h, k), where h is the x-coordinate and k is the y-coordinate. To find the vertex, we can use the formula: h = -b / (2a) and substitute the values of a, b, and c from the quadratic equation. In this case, a = -6, b = 5, and c = 2.

h = -5 / (2 * -6)
= -5 / -12
= 5 / 12

To find the y-coordinate of the vertex, substitute the x-coordinate (h) back into the equation: f(x) = -6x^2 + 5x + 2.
k = f(h)
= -6(5/12)^2 + 5(5/12) + 2
= -6(25/144) + 5(5/12) + 2
= -25/24 + 25/12 + 2
= -25/24 + 50/24 + 48/24
= 73/24

Therefore, the vertex is (5/12, 73/24).

2. Axis of Symmetry: The axis of symmetry is a vertical line that passes through the vertex of the quadratic function. The equation of the axis of symmetry is given by x = h, where h is the x-coordinate of the vertex. In this case, the axis of symmetry is x = 5/12.

3. Range: The range is the set of all possible y-values (outputs) of the function. For a quadratic function in the form f(x) = ax^2 + bx + c, where a < 0, the range is (-∞, k], where k is the y-coordinate of the vertex. In this case, the range is (-∞, 73/24].

4. Minimum or Maximum Value: Since the coefficient of x^2 is negative in the function f(x) = -6x^2 + 5x + 2, it is a downward-opening parabola, and the vertex represents the maximum value. The maximum value is k = 73/24.

5. Increasing and Decreasing Intervals: To find the intervals where the function is increasing or decreasing, we need to analyze the coefficient of x^2. Since it is negative, the function is decreasing on the interval (-∞, 5/12) and increasing on the interval (5/12, ∞).

6. Zeros: The zeros of the quadratic function f(x) = -6x^2 + 5x + 2 correspond to the values of x for which f(x) = 0. To find the zeros, we need to set f(x) equal to zero and solve for x.

-6x^2 + 5x + 2 = 0

We can either factor the quadratic equation or use the quadratic formula to solve for x. If factoring is not possible, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -6, b = 5, and c = 2. Substituting these values into the formula, we get:

x = (-5 ± √(5^2 - 4(-6)(2))) / (2(-6))
= (-5 ± √(25 + 48)) / (-12)
= (-5 ± √73) / (-12)

Therefore, the zeros are x = (-5 + √73) / (-12) and x = (-5 - √73) / (-12).

7. Standard Form: The standard form of a quadratic equation is f(x) = a(x - h)^2 + k, where (h, k) is the vertex. Substituting the values of the vertex (5/12, 73/24), we get:

f(x) = a(x - 5/12)^2 + 73/24

Therefore, the quadratic equation in standard form is f(x) = -6(x - 5/12)^2 + 73/24.