Question

Find the minimum value of the function f(x)=0.1x^2-x+8.1f(x)=0.1x 2 −x+8.1 to the nearest hundredth.

Answer 1

The minimum value of the function is 5.6.

Explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

`f(x) = ax^(2) + bx + c`

It's vertex is the point
`(x_(v), y_(v))`

In which

`x_(v) = -(b)/(2a)`

`y_(v) = -(\Delta)/(4a)`

Where

`\Delta = b^2-4ac`

If a<0, the vertex is a maximum point, that is, the maximum value happens at
`x_(v)`, and it's value is
`y_(v)`.

If a > 0, then
`y_(v)` is the minimum value of the function.

In this question:

We are given the following function:

`f(x) = 0.1x^2 - x + 8.1`

Which is a quadratic function with
`a = 0.1, b = -1, c = 8.1`.

To find the minimum value, we have that:

`\Delta = b^2-4ac = (-1)^2 - 4(0.1)(8.1) = -2.24`

`y_(v) = -(-2.24)/(4(0.1)) = 5.6`

The minimum value of the function is 5.6.