Question

find the zeros -4 and 4 and vertex at (0, -16)

Answer 1

you mean find the quadratic function which has zeroes .... right ?

Answer 2

yes the quadratic function with zeros -4 and 4 and vertex at 0, -16

Answer 3

use the formula :
y = a(x-a)(x-b)
with a,b are the zeroes (here known a=-4 and b=4)
put a,b into equation above :

y = a(x-a)(x-b)
y = a(x-(-4))(x-4)
y = a(x+4)(x-4)

now, subtitute the vertex point (0,-16), giving us
y = a(x+4)(x-4)
-16 = a(0+4)(0-4)
-16 = a(4)(-4)
-16 = -16a
thus a = -16/-16 = 1

therefore, it is y = a(x-a)(x-b) = 1(x+4)(x-4) = (x+4)(x-4) = x^2 - 16

Answer 4

i am trying to answer this problem and the options are y=x squared -16, y= -x squared -16, y= x squared +16, or y= -x squared +16

Answer 5

so the answer is a i see know how you came up with that thank you very much.

Answer 6

yup
welcome :)

Answer 7

how do you find the domain of y= 1/x-5 +7