How do you graph the function, label the vertex, axis of symmetry, and xintercepts. #y=x^216x + 63#?
1 Answer
May 7, 2015

Has a yintercept at
#y=63# as determined by setting#x=0# (and, yes; I know this wasn't asked for but it might help in graphing) 
Since
#(x^216x+63)# can be factored as
#(x7)(x9)# , the xintercepts are at#7# and#9# 
The vertex occurs at the point where
#y'=0#
#y'=2x16= 0 rarr x=8# (This could also be determined as the mid x coordinate between the two xintercepts).
At#x=8#
#y=8^216(8)+63 = 0#
So the vertex is at#(8,0)# 
This equation is that of a standard parabola so the axis of symmetry is the vertical line through the vertex:
#x=8#
graph{x^216x+63 [5.47, 22.99, 2, 12.25]}
I'm not very good at drawing smooth curves so I've used an external tool; the labeling is up to you)