Question

Marsha wants to determine the vertex of the quadratics vertex of the quadratics functions f(x)=x^2-x+2

Answer 1

The vertex of the function is
`((1)/(2), (7)/(4))`

Explanation:

Suppose we have a quadratic equation in the following format:

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

The vertex of the function is the point
`(x_(v), f(x_(v))`, in which

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

In this problem:

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

This means that
`a = 1, b = -1, c = 2`

So

`x_(v) = -(-1)/(2) = (1)/(2)`

`f(x_(v)) = f((1)/(2)) = ((1)/(2))^(2) - (1)/(2) + 2 = (1)/(4) - (1)/(2) + 2 = (1 - 2 + 8)/(4) = (7)/(4)`

The vertex of the function is
`((1)/(2), (7)/(4))`