Question 392180
  <pre><font size = 3 color = "indigo"><b>
Hi

p(x)=x^2-2x+5 
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
completing the square to put into vertex form
p(x) = (x-1)^2 + 4  vertex Pt(1,4) is a minimum pt, a>0, parabola opens upward
Although the range of p(x) is p(x)>= 4, the domain x-values are all real numbers
{{{drawing(300,300, -10,10,-10,10,blue(line(1,10,1,-10))   
 grid(1),
circle(1, 4,0.4),
graph( 300, 300, -10,10,-10,10,x^2-2x+5
))}}}