\( 1 \leftarrow \quad \) Identify the vertex, axis of symmetry, \( y \)-intercept, \( x \)-intercepts, and opening of the parabola, then sketch the graph.

We start with the quadratic in standard form:   y = x² – 2x + 1 Step 1. Identify the vertex.  The vertex of a quadratic y = ax² + bx + c is found using the formula for the x-coordinate:   x = –b/(2a) For this equation, a = 1 and b = –2, so:   x = –(–2)/(2·1) = 2/2 = 1 Now substitute x = 1 back into the equation:   y = (1)² – 2(1) + 1 = 1 – 2 + 1 = 0 Thus, the vertex is (1, 0). Step 2. Find the axis of symmetry.  The axis of symmetry is the vertical line that passes through the vertex. Therefore:   Axis of symmetry: x = 1 Step 3. Find the y-intercept.  The y-intercept is found by setting x = 0:   y = (0)² – 2(0) + 1 = 1 So, the y-intercept is at (0, 1). Step 4. Find the x-intercepts.  The x-intercepts are found by setting y = 0:   0 = x² – 2x + 1   0 = (x – 1)² This gives a double root:   x – 1 = 0 ⟹ x = 1 Thus, the only x-intercept is at (1, 0). Step 5. Determine the direction of opening.  Since the coefficient a = 1 is positive, the parabola opens upward. Step 6. Sketch the graph. 1. Plot the vertex at (1, 0). 2. Draw the axis of symmetry, the vertical line x = 1. 3. Mark the y-intercept, (0, 1). By symmetry, there will be a corresponding point on the opposite side of the axis: (2, 1). 4. Note that the parabola touches the x-axis only at (1, 0). 5. Draw a smooth curve opening upward that passes through these points. Below is a rough sketch: ↑ 2 | • (2,1) 1 | • (0,1) | | 0 |--------•--------→ | (1,0) (vertex and x-intercept) -1 | | +---------------------------→ 0 1 2 3 Summary:  • Vertex: (1, 0)  • Axis of symmetry: x = 1  • y-intercept: (0, 1)  • x-intercept: (1, 0) (double root)  • Opening: Upward This completes the identification and sketch of the graph of y = x² – 2x + 1.