1) \( f(x)=x^{2}-x+1 \)

The function you provided is a quadratic function given by: \[ f(x) = x^2 - x + 1 \] ### Properties of the Function 1. **Vertex**: The vertex of a quadratic function in the form \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -1 \). \[ x = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \] To find the y-coordinate of the vertex, substitute \( x = \frac{1}{2} \) back into the function: \[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \] Thus, the vertex is at \( \left(\frac{1}{2}, \frac{3}{4}\right) \). 2. **Axis of Symmetry**: The axis of symmetry is the vertical line that passes through the vertex. For this function, it is: \[ x = \frac{1}{2} \] 3. **Y-intercept**: The y-intercept occurs when \( x = 0 \): \[ f(0) = 0^2 - 0 + 1 = 1 \] So, the y-intercept is at \( (0, 1) \). 4. **X-intercepts**: To find the x-intercepts, set \( f(x) = 0 \): \[ x^2 - x + 1 = 0 \] The discriminant \( D \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \): \[ D = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \] Since the discriminant is negative, there are no real x-intercepts. The function does not cross the x-axis. 5. **Graph**: The graph of the function is a parabola that opens upwards (since \( a > 0 \)), with its vertex at \( \left(\frac{1}{2}, \frac{3}{4}\right) \) and a y-intercept at \( (0, 1) \). The function is always positive for all real \( x \). ### Summary - Vertex: \( \left(\frac{1}{2}, \frac{3}{4}\right) \) - Axis of symmetry: \( x = \frac{1}{2} \) - Y-intercept: \( (0, 1) \) - No real x-intercepts (the function does not cross the x-axis). - The function is always positive. If you have any specific questions or need further analysis, feel free to ask!