Question 963255
Complete the square to convert to vertex form,
{{{y=x^2-6x+10}}}
{{{y=(x^2-6x+9)+10-9}}}
{{{y=(x-3)^2+1}}}
The vertex occurs at ({{{3}}},{{{1}}}).
The parabola opens upwards. 
The interval in which it is decreasing is ({{{-infinity}}},{{{3}}}).
The interval in which it is increasing is ({{{3}}},{{{infinity}}}).
The axis of symmetry is {{{x=3}}}.
The parabola opens upwards so the vertex value is the minimum.
The range is [{{{1}}},{{{infinity}}}).
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*[illustration para2.JPG].