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

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Answer 1 · 2015-05-27T00:07:16

to find x-intercept , make y = 0 --> ${(x - 3)}^{2} = \frac{- 16}{- 4} = 4$ $(x - 3)^2 = (-16)/-4 = 4$

x1 - 3 = 2 -> x1 = 5

x2 = 3 - 2 ..-> x2 = 1

To find y-intercept, make x = 0 -> $y = - 36 + 16 = - 20$ $y = -36 + 16 = -20$

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