How do you graph y = x^2 - 2x + 1?

To graph \( y = x^2 - 2x + 1 \), plot the vertex, axis of symmetry, and a few points on either side.

First, identify the vertex of the parabola. The given equation is in the form \( y = ax^2 + bx + c \). Here, \( a = 1 \), \( b = -2 \), and \( c = 1 \). The vertex can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values, we get \( x = -\frac{-2}{2 \cdot 1} = 1 \). To find the y-coordinate of the vertex, substitute \( x = 1 \) back into the equation: \( y = (1)^2 - 2(1) + 1 = 0 \). So, the vertex is at \( (1, 0) \).

Next, determine the axis of symmetry, which is the vertical line that passes through the vertex. For this equation, the axis of symmetry is \( x = 1 \).

Now, choose a few x-values on either side of the vertex to find corresponding y-values. For example, if \( x = 0 \), then \( y = (0)^2 - 2(0) + 1 = 1 \). If \( x = 2 \), then \( y = (2)^2 - 2(2) + 1 = 1 \). Plot these points: \( (0, 1) \) and \( (2, 1) \).

To get a more accurate graph, you can plot additional points. For instance, if \( x = -1 \), then \( y = (-1)^2 - 2(-1) + 1 = 4 \). If \( x = 3 \), then \( y = (3)^2 - 2(3) + 1 = 4 \). Plot these points: \( (-1, 4) \) and \( (3, 4) \).

Finally, draw a smooth curve through all the points, ensuring it is symmetrical about the axis of symmetry. This will give you the graph of \( y = x^2 - 2x + 1 \).