Question

i need to write the vertex form of the following parabola: 1/2(y+4)=(x-7)^2

Answer 1

http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php

Answer 2

put x-7=0
and y+4=0
solve and you get the vertex.

Answer 3

\[{1\over2}(y+4)=(x-7)^2\]\[y+4=2(x-7)^2\]\[y=2(x-7)^2-4\]

Answer 4

what you get same thing
from my calculation x=7
y=-4
vertex (7,-4)

Answer 5

yes but why would you do that when the vertex is given in the equation..

Answer 6

the form it was in:\[{1\over a}(y-k)=(x-h)^2\]where \((h,k)\) is the vertex
the form we need is the standard form:\[y=a(x-h)^2+k\]you literally just rearrange it :)