How do you sketch the graph of #y=x^2-2x# and describe the transformation?

#color(magenta)("Preamble")#

As the coefficient of #x^2# is positive #(+1x^2)# then the graph is of form #uu#. Thus the vertex is a minimum.

#color(red)("If")# the coefficient had been negative then the graph would have been in the form #nn#. Thus the vertex would have been a maximum.
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#color(magenta)("Answering the question")#

What we are transforming is the basis of #y=x^2# where #x_("vertex")=0#
Let the vertex of #y=x^2->(x_1,y_1)=(0,0)#

Note this is the same as: #y=x^2+0x+0#
Note that the y-intercept is at #x=0#
So for this case the y-intercept is #y=(0)^2+0x+0=0#

#color(blue)("Transformation left or right - Shift left or right")#

Let the vertex of #y=x^2-2x ->(x_2,y_2)#

By including the #color(red)(-2)color(green)(x)# the new #x_("vertex")# of #color(green)(y=x^2color(red)(-2)x)# is #(-1/2)xxcolor(red)(-2)=+1 =x_2#

So the transformation for #x# is #x_2-x_1" "=" "1-0=+1#

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#color(blue)("Transformation up or down - shift up or down")#

#y_(vertex)" for "y=x^2=0=y_1#

The new #y_("vertex") =y_2 # at #x_2=1#

So by substitution for #x# #y_2=(x_2)^2-2(x_2)" "=" "(1)^2-2(1)=-1#

Thus transformation for y is #y_2-y_1" "=" "-1-0" "=" "-1#
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Consequently the transformation of any point is #(x_1+1,y_1-1)#