How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=x^2-4x+4#?

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Answer 1 · 2016-11-13T17:17:38

#y#-intercept is #4#, axis of symmetry is #x-2=0# and vertex is #(2,0)#

#y#-intercept is given by #y=f(0)=0^2-4xx0+4=4#,

hence, #y#-intercept is #4#.

Now #y=f(x)=x^2-4x+4#

= #(x-2)^2#

Hence, axis of symmetry is #x-2=0#

and at #x=2#, #y=f(x)=0#

Hence, vertex is #(2,0)#
graph{x^2-4x+4 [-7.625, 12.375, -2.12, 7.88]}