How do you graph # y = (x-2)^2 +2#?

If your question wording 'how do you graph' were to be taken literally you would get the following answer:

Draw up a table of values by substituting for #x# and calculate the related values of y. Mark off the points and draw the best fit that you can where the line passes through each marked point.
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Assumption: You intend the question to be: Determine the important points of #y=(x-2)^2+2# and sketch the graph.

Let any point on the graph be #P#

#color(blue)("Determine the general shape of the graph")#

Expanding the brackets of #y=+1(x-2)^2+2# we have:

#y=+x^2-4x+6#

As the coefficient of #x^2# is positive we have the general shape of #uu#. Thus the vertex is a minimum.

Note that if the coefficient had been negative then you would have had the general shape of #nn#

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#color(blue)("Determine the vertex")#

The given equation format is that of 'vertex form'

From this we can directly read off the vertex coordinates:

#color(blue)(P_("vertex")->(x,y)=(2,2))#
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#color(blue)("Determine the y-intercept")#

Substitute #x=0#

#y_("intercept")=(0-2)^2+2 = 6#

#color(blue)("P_("y-Intercept") ->(x,y)=(0,6)"#
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#color(blue)("Determine the x-intercept")#

Set #y=0#

#0=(x-2)^2+2#

Subtract 2 from both sides

#(x-2)^2=-2#

Square root both sides

#x-2=+-sqrt(-2)#

The square root of a negative number means that the graph does not cross the x-axis.

#color(brown)("If you required to have some actual values for x-intercept")#

#x=+2+-sqrt(2xx(-1))#

#x=+2+-sqrt(2)sqrt(-1)#

#x=+2+-sqrt(2)color(white)(.)i#

Where this is of complex number type of format:

'Real number' + 'Imaginary number' #-> Re+ Im#