How do you find the zeros of #y=-x^2+4x-4#?

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Answer 1 · 2014-11-06T18:09:32

By setting it equal to zero,

#-x^2+4x-4=0#

by multiplying by #-1#,

#=> x^2-4x+4=0#,

by factoring out,

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

Hence, #x=2# is the zero of the function.

I hope that this was helpful.