3 Homework Question 9, 4.3.51 HW Score: Part 1 of 11 Points: For the quadratic function $f(x)=2 x^{2}-x+1$, answer parts (a) through (f). (a) Find the vertex and the axis of symmetry of the quadratic function, and determine whether the graph is concave up or concave down. The vertex is (Simplify your answer. Type an ordered pair, using integers or fractions.)

Step 1 ：The vertex of a quadratic function in the form \(f(x) = ax^2 + bx + c\) is given by the point \((-\frac{b}{2a}, f(-\frac{b}{2a}))\). The axis of symmetry is the vertical line \(x = -\frac{b}{2a}\). The graph is concave up if \(a > 0\) and concave down if \(a < 0\).

Step 2 ：In this case, \(a = 2\), \(b = -1\), and \(c = 1\). So, the vertex is \((-\frac{-1}{2*2}, f(-\frac{-1}{2*2}))\) and the axis of symmetry is \(x = -\frac{-1}{2*2}\). Since \(a = 2 > 0\), the graph is concave up.

Step 3 ：Let's calculate the coordinates of the vertex and the equation of the axis of symmetry.

Step 4 ：\[a = 2\]

Step 5 ：\[b = -1\]

Step 6 ：\[c = 1\]

Step 7 ：\[vertex_x = 0.25\]

Step 8 ：\[vertex_y = 0.875\]

Step 9 ：\[axis_of_symmetry = 0.25\]

Step 10 ：Final Answer: The vertex of the quadratic function is \(\boxed{(0.25, 0.875)}\), the axis of symmetry is \(\boxed{x = 0.25}\), and the graph is concave up.