Given:
![x^2-2x+y-4=0](https://img.qammunity.org/2023/formulas/mathematics/college/ix1f7dtvvq6q2b3d3dpynj69vh5w7xro18.png)
Let's complete the square to find the vertex of the parabola.
To solve first move all terms not containing y to the right side of the equation:
![y=-x^2+2x+4](https://img.qammunity.org/2023/formulas/mathematics/college/49nmelpta0yj6mrprh0jfuv8jgw7facmji.png)
Now, take the vertex form of a parabola:
![y=a(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/college/97p0xsjs0cwme4ddvwkim2cbbqprhnlhsv.png)
Apply the standard form of a parabola:
![\begin{gathered} ax^2+bx+c \\ \\ -x^2+2x+4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lw0nt6leulrjmau2c2y3lt3qjyb74b28wh.png)
Thus, we have:
a = -1
b = 2
c = 4
Now, to find the value of h, we have:
![\begin{gathered} h=-(b)/(2a) \\ \\ h=-(2)/(2(-1)) \\ \\ h=-(2)/(-2) \\ \\ h=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sbe76vbhj968nitmwie27ylo2egqex8xfj.png)
To find the value of k, we have:
![\begin{gathered} k=c-(b^2)/(4a) \\ \\ k=4-(2^2)/(4(-1)) \\ \\ k=4-(4)/(-4) \\ \\ k=4+1 \\ \\ k=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qe0nhgzx9tf42remuaxkz7qzle9z7heiit.png)
We have the values:
h = 1
k = 5
The vertex of the parabola is:
(h, k) ==> (1, 5)
ANSWER:
(1, 5)