Answer:
The vertex of the function is [tex](\frac{1}{2}, \frac{7}{4})[/tex]
Step-by-step explanation:
Suppose we have a quadratic equation in the following format:
[tex]y = f(x) = ax^{2} + bx + c[/tex]
The vertex of the function is the point [tex](x_{v}, f(x_{v})[/tex], in which
[tex]x_{v} = -\frac{b}{2a}[/tex]
In this problem:
[tex]f(x) = x^{2} - x + 2[/tex]
This means that [tex]a = 1, b = -1, c = 2[/tex]
So
[tex]x_{v} = -\frac{-1}{2} = \frac{1}{2}[/tex]
[tex]f(x_{v}) = f(\frac{1}{2}) = (\frac{1}{2})^{2} - \frac{1}{2} + 2 = \frac{1}{4} - \frac{1}{2} + 2 = \frac{1 - 2 + 8}{4} = \frac{7}{4}[/tex]
The vertex of the function is [tex](\frac{1}{2}, \frac{7}{4})[/tex]
