Question 246225: sketch the graph of f(x)=(x-3)^2-1 identifying the vertex, axis of symmetry and all intercepts. Must find at least 5 points.
i have gotten this far i think, not even 50% sure
x f(x)=(x-3)^2-1
-2 =(-2-3)^2-1=(-5)^2-1=24 (-2,24)
-1 =(-1-3)^2-1=(-4)^2-1=15 (-1,15)
0 =(0-3)^2-1=(-3)^2-1=8 (0,8)
1 =(1-3)^2-1=(-2)^2-1=3 (1,3)
2 =(2-3)^2-1=(-1)^2-1=0 (2,0)
am i correct in the points?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the squared term is always positive; so the minimum for f(x) is -1 (when the squared term is zero @ x=3)
___ this is the axis of symmetry (which contains the vertex (3,-1))
___ your points should be around x=3 to show the desired details
___ you have two of the intercepts (one x and one y) and you need to show the third (second x)
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