What is the vertex form of #y = x^2 -14x + 16#?

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Answer 1 · 2018-03-14T18:15:45

#y=(x-7)^2-33#

First find the vertex using the formula
#x=(-b)/"2a"#

a=1
b=-14
c=16

#x=(-(-14))/"2(1)"# This simplifies to #x=14/"2"# which is 7.
so #x=7#

So on now that we have x we can find y.

#y=x^2-14x+16#
#y=(7)^2-14(7)+16#
#y=-33#

#Vertex = (7,-33)# where h=7 and k=-33

We now finally enter this into vertex form which is,
#y=a(x-h)^2+k#

x and y in the "vertex form" are not associated with the values we found earlier.

#y=1(x-7)^2+(-33)#
#y=(x-7)^2-33#