f(x) = 6x^2 + 7x − 20
Find the maximum or minimum value of f(x). (Round your answer to two decimal places.)
To find the maximum or minimum value of a quadratic function, you need to determine the vertex of the parabola. For a quadratic function in the form f(x) = ax^2 + bx + c, the x-coordinate of the vertex can be found using the formula x = -b / (2a). Once you have the x-coordinate, you can substitute it back into the function to find the y-coordinate, which gives the maximum or minimum value of f(x).
In the given function, f(x) = 6x^2 + 7x − 20, we can observe that a = 6, b = 7, and c = -20.
Using the formula x = -b / (2a), we have:
x = -(7) / (2 * 6)
x = -7 / 12
Now substitute this value back into the function to find the y-coordinate:
f(-7/12) = 6(-7/12)^2 + 7(-7/12) − 20
To simplify the calculation, simplify the fraction (-7/12)^2 to (49/144):
f(-7/12) = 6(49/144) + 7(-7/12) − 20
Multiply the fractions and simplify:
f(-7/12) = 294/144 - 49/12 - 20
To add the fractions, find a common denominator, which is 144:
f(-7/12) = 294/144 - 588/144 - 2880/144
f(-7/12) = -3174/144
Now convert the fraction to a decimal by dividing the numerator by the denominator:
f(-7/12) ≈ -22.08
Hence, the maximum or minimum value of f(x) is approximately -22.08.
recall that the vertex of a parabola is at x = -b/2a = -7/12
So evaluate f(x) there.
I'm sure you can tell whether it is a max or min.