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May 8th, 2017, 06:14 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, 07:01 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
vertex form, $y=a(xh)^2+k$, 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, 04:34 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 (xh)^4 + K how does this equate in a quartic form:y = ax4 + (????)x + (????)x ..... etc thanks 
May 21st, 2017, 05:03 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
Use the binomial theorem to expand $(xh)^4$ then distribute $a$ ...


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