Answer:
The minimum value of the function is 5.6.
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
If a > 0, then [tex]y_{v}[/tex] is the minimum value of the function.
In this question:
We are given the following function:
[tex]f(x) = 0.1x^2 - x + 8.1[/tex]
Which is a quadratic function with [tex]a = 0.1, b = -1, c = 8.1[/tex].
To find the minimum value, we have that:
[tex]\Delta = b^2-4ac = (-1)^2 - 4(0.1)(8.1) = -2.24[/tex]
[tex]y_{v} = -\frac{-2.24}{4(0.1)} = 5.6[/tex]
The minimum value of the function is 5.6.
