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Find the vertex f(x)= 3x^2-18x+3 please help me. Awards will be given

Mathematics
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use the all mighty formula \[x= -\frac{b}{2a}\] to find the first coordinate
in this case \[a=3,b=-18\] so \[-\frac{b}{2a}=-\frac{-18}{2\times3}=3\]
that is the first coordinate and the second coordinate is what you get when you replace x by 3. you get \[3\times 3^2-18\times 3+3=27-54+3=-27+3=-24\] check my arithmetic because it is late

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also because i didn't really compute. as soon as i saw 27 i knew the next one would be -54. always works that way
vertex is (3,-24) unless i messed up
so (3,-24) ? and that would make it a maximum right?
heck no
this is a parabola that faces up. no max, but a min
|dw:1314675140476:dw|
the minimum value is -24
Ah I got it mixed up!
I always do that with the minimum maximum.

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