Find the minimum value of the function f(x) = 0.8x2 + 11.2x + 42 to the
nearest hundredth.

The minimum value is 2.8.  The vertex is (-7, 2.8).

Step-by-step explanation:

Through the x^2 term we identify this as a quadratic function whose graph opens up.  The minimum value of this function occurs at the vertex (h, k).  The x-coordinate of the vertex is

        -b                                                    11.2

x =             which here is x = - = -7

         2a                                                 2(0.8)

We need to calculate the y value of this function at x = -7.  This y-value will be the desired minimum value of the function.

Substituting -7 for x in the function f(x) = 0.8x^2 + 11.2x + 42 yields

f(-7) = 0.8(-7)^2 + 11.2(-7) + 42 = 2.8

The minimum value is 2.8.  The vertex is (-7, 2.8).

the answer is 4.8 units on e2020.