# How do we find the zeros from y=-4(x-3)^2+16 vertex form equation?

to find x-intercept , make y = 0 -->${\left(x - 3\right)}^{2} = \frac{- 16}{-} 4 = 4$
To find y-intercept, make x = 0 -> $y = - 36 + 16 = - 20$