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

Nov 6, 2014

By setting it equal to zero,

$- {x}^{2} + 4 x - 4 = 0$

by multiplying by $- 1$,

$\implies {x}^{2} - 4 x + 4 = 0$,

by factoring out,

$\implies {\left(x - 2\right)}^{2} = 0$

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

I hope that this was helpful.