Consider the standard form of a quadratic equation is,
[tex]y=ax^2+bx+c[/tex]It's vertex (h,k) is given by,
[tex]\begin{gathered} h=\frac{-b}{2a} \\ k=f(\frac{-b}{2a}) \end{gathered}[/tex]Now consider the given equation,
[tex]y=x^2-2x+4[/tex]So its vertex will be,
[tex]\begin{gathered} h=\frac{-(-2)}{2(1)}=1 \\ k=(1)^2-2(1)+4=1-2+4=3 \end{gathered}[/tex]So the vertex of the quadratic equation lies at (1,3).
Now, observe the given graphs.
It is found that only first graph shows the vertex at point (1,3).
So option a will be the correct choice.
