Find the extreme values of #y= x^2 − 2x + 3#?

To get the minimum set the derivative #d/dx(x^2-2x+3)=2x-2#
equal to zero:
#2x-2=0#
which solves for #x=1#
Since the highest exponent in this equation is 1 it only has one solution.
By doing the second derivative #d/dx(2x-2)=2# you can see at x=1 its positive therby its a minimum.
The maximums are, since the highest exponent in the original equation (2) is even and has a positive factor, at #+oo# and #-oo#

The term #x^2# 'grows' faster than the other term involving #x#

So to use a none mathematical term; #x^2# wins.

So #lim_(x->oo) y=[lim_(x->oo)x^2]-[2lim_(x->oo)x]+3 #

Tends to

#lim_(x->oo) y=[lim_(x->oo)x^2]#

Note that if #x<0# then #x^2>0#
Also that if #x>0# then #x^2>0# as well

Thus we write:

#lim_(x->oo) y=[lim_(x->oo)x^2] -> k# where #k=+oo#
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Foot note

As the #x^2# term is positive then the general shape is that of type #uu#. So the #+oo # matches that context.

Given: #y=x^2-2x+3#

#x # can and may take on the value of #0#

As the #x^2# term is positive then the general shape of the curve is that of #uu#. Thus there is a minimum for #y#

To determine this we need the vertex (bottom of the #uu#)

This will be #1/2# way between the x-intercepts if there are any.
Looking at the given equation notice that #(-1)xx(-3)=+3#

But #-1-3 !=-2# thus it is more straight forward to use the formula or you can do this sort of cheat. It is part of the various steps to complete the square.

Given
#y=ax^2+bx+c# write as #y=a(x+b/(2a))^2+k#

#x_("vertex") =(-1)xxb/(2a) = (-1) xx (-2)/(2xx1)=+1#

Substitute #x=1# into the original equation.

#y_("vertex")=(1)^2-2(1)+3 = +2#

Vertex #->(x,y)=(1,2)#