My Math Forum Convert vertex form to standard form

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 May 8th, 2017, 05:14 AM #1 Newbie   Joined: May 2017 From: Australia Posts: 2 Thanks: 0 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
 May 8th, 2017, 06:01 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,518 Thanks: 1240 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.
 May 21st, 2017, 03:34 AM #3 Newbie   Joined: May 2017 From: Australia Posts: 2 Thanks: 0 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
 May 21st, 2017, 04:03 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,518 Thanks: 1240 Use the binomial theorem to expand $(x-h)^4$ then distribute $a$ ...

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