Posted by -Untamed- on Monday, December 5, 2011 at 8:08pm.
The canonical form of a quadratic equation is:
y=a(x-h)²+k
where
a is a constant
(h,k) is the vertex.
So in the given case, h=3, k=-4, and a remains to be found.
We can find a since we know the curves passes through (4,1), hence
1=a(4-3)²-4
1=a-4
solve for a to get
a=5
Check:
y=5(x-3)²-4
For x=4, y=5(1)²-4=1, OK
For the general form, expand the equation
y=5(x-3)²-4
y=5x²-30x+41
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