1. The problem is to graph the function $y = x^2 - 2x + 1$ and understand its shape. 2. The function is a quadratic polynomial, which graphs as a parabola. The general form is $y = ax^2 + bx + c$. 3. Here, $a = 1$, $b = -2$, and $c = 1$. Since $a > 0$, the parabola opens upwards. 4. To find the vertex, use the formula for the x-coordinate: $x = -\frac{b}{2a} = -\frac{-2}{2 \times 1} = 1$. 5. Substitute $x=1$ into the function to find the y-coordinate: $y = 1^2 - 2 \times 1 + 1 = 1 - 2 + 1 = 0$. 6. So, the vertex is at $(1, 0)$, which is also the minimum point of the parabola. 7. The function can be factored as $y = (x - 1)^2$, showing it touches the x-axis at $x=1$. 8. The parabola is symmetric about the vertical line $x=1$. 9. The horizontal line $y=0$ is the x-axis, which the parabola touches at the vertex. 10. Plotting these shows the parabola opening upwards with its lowest point on the x-axis at $(1,0)$, and the x-axis as a horizontal line. Final answer: The graph of $y = x^2 - 2x + 1$ is a parabola opening upwards with vertex at $(1,0)$, touching the x-axis there.