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

Question

13042 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,

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

by multiplying by $- 1$ $-1$ ,

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

by factoring out,

$\Rightarrow {(x - 2)}^{2} = 0$ $=> (x-2)^2=0$

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

I hope that this was helpful.

How do you find the zeros of y=−x2+4x−4y=-x^2+4x-4?