To find the vertex of the quadratic function \( f(x) = x^2 - x + 2 \), we can use the vertex formula for a quadratic function in the form \( f(x) = ax^2 + bx + c \). The x-coordinate of the vertex is given by:
\[ x = -\frac{b}{2a} \]
Here, \( a = 1 \) and \( b = -1 \).
Substituting the values into the formula:
\[ x = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]
Now, we need to find the y-coordinate of the vertex by substituting \( x = \frac{1}{2} \) back into the function \( f(x) \):
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 2 \]
Calculating this step-by-step:
- \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
- \( -\left(\frac{1}{2}\right) = -\frac{1}{2} = -\frac{2}{4} \) (convert to have a common denominator)
- Adding these: \[ f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} + 2 \]
- \( 2 = \frac{8}{4} \) (convert to have a common denominator)
- Now calculating: \[ f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} = \frac{1 - 2 + 8}{4} = \frac{7}{4} \]
Thus, the vertex of the quadratic function \( f(x) = x^2 - x + 2 \) is
\[ \left(\frac{1}{2}, \frac{7}{4}\right) \]
Therefore, the correct answer is:
(one-half, seven-quarters)