The equation:
![y=(x-1)^2+2](https://img.qammunity.org/2023/formulas/mathematics/college/eajr1e109lce1hu2ocl5ccyza5baso4l6x.png)
has the form:
![y=(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/high-school/h4kqjwwshdhi39of5qh206bvx52ti58zvg.png)
where the point (h, k) is the vertex of the parabola. In this case, the vertex is located at (1, 2).
The axis of symmetry is
x = h
x = 1
To graph the function we need three points. One of them is the vertex. The other two should be at the same distance from the axis of symmetry. Substituting with x = 0 into the equation, we get:
![\begin{gathered} y=(0-1)^2+2 \\ y=1+2 \\ y=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/214rrt07k36ti5atjuxwu2t1ial8r2xwxs.png)
Substituting with x = 2 into the equation, we get:
![\begin{gathered} y=(2-1)^2+2 \\ y=1+2 \\ y=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/t5utom5qazxle34v1h4yfq57ffu39oa7ff.png)
Connecting with a parabola the points (0, 3), (1, 2), and (2, 3) we get: