Given:
[tex]x^2-2x+y-4=0[/tex]
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:
[tex]y=-x^2+2x+4[/tex]
Now, take the vertex form of a parabola:
[tex]y=a(x-h)^2+k[/tex]
Apply the standard form of a parabola:
[tex]\begin{gathered} ax^2+bx+c \\ \\ -x^2+2x+4 \end{gathered}[/tex]
Thus, we have:
a = -1
b = 2
c = 4
Now, to find the value of h, we have:
[tex]\begin{gathered} h=-\frac{b}{2a} \\ \\ h=-\frac{2}{2(-1)} \\ \\ h=-\frac{2}{-2} \\ \\ h=1 \end{gathered}[/tex]
To find the value of k, we have:
[tex]\begin{gathered} k=c-\frac{b^2}{4a} \\ \\ k=4-\frac{2^2}{4(-1)} \\ \\ k=4-\frac{4}{-4} \\ \\ k=4+1 \\ \\ k=5 \end{gathered}[/tex]
We have the values:
h = 1
k = 5
The vertex of the parabola is:
(h, k) ==> (1, 5)
ANSWER:
(1, 5)
