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May 8th, 2017, 06:14 AM   #1
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Convert vertex form to standard form

Hi, I have a parabola with the following vertex info, I am trying to convert this to standard form (ax2 + bx + c), however I cannot seem to get the conversion correct. I have plots for the parabola and have created a trend line with equation in excel so I can compare and be confident my conversion is correct: Please see below, I have followed the rules for conversion but something is incorrect, if anyone could help it would be greatly appreciated and show the working out.
y = -a (x - h)^2 + K
K = 114, h = 0, x = -255, y = 0
from this info I have derived a = -0.00175

from my excel trendline the standard form should be approx =
y = -0.0018x^2 + 0.0027x + 116.44

thanks
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May 8th, 2017, 07:01 AM   #2
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vertex form, $y=a(x-h)^2+k$, expanded to standard form, $y=ax^2+bx+c$ yields ...

$y=ax^2 + (-2ah)x + (ah^2+k)$

note $b=-2ah$ ... if $h=0$, then there is no linear term.
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May 21st, 2017, 04:34 AM   #3
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thanks skeeter, thats what I wanted. How do I convert a quartic form to vertex,
so y = a (x-h)^4 + K

how does this equate in a quartic form:y = ax4 + (????)x + (????)x ..... etc

thanks
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May 21st, 2017, 05:03 AM   #4
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Use the binomial theorem to expand $(x-h)^4$ then distribute $a$ ...
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