How do you use the important points to sketch the graph of #y=(x-2)^2#?

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Answer 1 · 2018-05-18T02:03:37

See below

#y=(x-2)^2#

#= x^2-4x+4#

Notice that #y# is a quadratic function of the form #ax^2+bx+c#, therefore its graph will be a parabola.

#y =0# at #x=2# - Notice that this is a coincident root meaning that #y# will touch the #x-#axis at #x=2# and at nowhere else.

Since #a =+1>0, y# will have a minimum value at
#x=(-b)/(2a) = 4/2 =2#

#:. y_min = 0# at #x=2#

Hence an "important point" is #(2,0)# being both the only #x-#intercept and the minimum of #y#

Next, we can find the #y-#intercept, where #x=0#

#y(0) = (-2)^2 =4#

So another "important point" is #(0,4)# the #x-#intercept

We can see these points on the graph of #y# below.

graph{(x-2)^2 [-6.83, 10.945, -1.68, 7.21]}