How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. #y=x^2-16x + 63#?
1 Answer
May 7, 2015
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Has a y-intercept 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^2-16x+63)# can be factored as
#(x-7)(x-9)# , the x-intercepts are at#7# and#9# -
The vertex occurs at the point where
#y'=0#
#y'=2x-16= 0 rarr x=8# (This could also be determined as the mid x coordinate between the two x-intercepts).
At#x=8#
#y=8^2-16(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^2-16x+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)
