Answer the question below about the quadratic function. f(x)=3x^2-30x+73 what is the function’s minimum or maximum value.

Answer 1

This question is asking you to find the vertex of the parabola since it will be the highest or lowest point of the quadratic equation. First, we will calculate the x-value at the vertex using a formula, then we will plug the x-value into the quadratic equation to find the y-value of the vertex.

To find the x-value of the vertex of a quadratic function, use the formula:

-b / 2a

where b is the constant by x, and a is the constant by x² . This is a standard formula. In this case:

a = 3

b = -30 (don't forget the negative sign!)

Now we can plug these into the formula to find the x-value of the vertex:

x-value = -b / 2a

x-value = -(-30) / (2 x 3)

x-value = 30 / 6 = 5

The x-value at the vertex is 5. To find the y-value at the vertex, plug the x-value we just found into the quadratic function:

f(x) = 3x² - 30x + 73

f(5) = 3(5)² - 30(5) + 73 = -2

The y-value at the vertex is -2, so our vertex is at (5,-2). Since the y-value at the vertex will always be the minimum or maximum value of a quadratic equation, -2 is the minimum or maximum value.

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Answer 2

The derivative is 6x-30 and = 0 at x=5.

The value at x=5: you should be able to verify that it is -2 and that it is a minimum.