How do you identify the important parts of #y=(x-2)^2# to graph it?

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Answer 1 · 2015-10-02T01:02:50

Vertex is #(2,0)#
Axis of symmetry #x=2#

#y=(x-2)^2#

It is a quadratic function in vertex form
#y = a(x-h)+k#
Where -
#(h,k)# is vertex
#x=h# is axis of symmetry.

In our case there is no #k# term. We shall have it as #0#

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

Co-ordinates of the vertex

#x=-1(h) = -1(-2)=2#
#y=k=0#

Vertex is #(2,0)#
Axis of symmetry #x=2#

Since #a# is positive, the curve is concave upwards.
It has a minimum.