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

Question

12010 views around the world

You can reuse this answer
Creative Commons License

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.