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November 13th, 2016, 06:42 AM  #1 
Newbie Joined: Sep 2016 From: zambia Posts: 27 Thanks: 0  Quadratic functions
The function f(x)=ax^2+bx+c has a maximum value of 4 where x=1.Find the value of a and b. after l know that the maximum point occurs at a(x+b/2a)^2+4c/4a(b^2/4a^2) hence the x value of the point is given by x=b/2a and y=(4c/4a)(b/4a)^2 substituting for x in x=b/2a 1=b/2a__________________________(i) and for y 4=(4c/4a)(b/4a)^2 _________________(ii) my problem is, how do you manipulate for the constant c so that in the two equations you only remain with a and b? 
November 13th, 2016, 07:53 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 479 Thanks: 203 
Yes, the maximum value of the parobola $ax^2 + bx + c$ occurs when $x = \ \dfrac{b}{2a}.$ But you are told that for this case $x = \ 1 \implies \ 1 = \ \dfrac{b}{2a} \implies b = 2a.$ And the maximum value of the parabola is $a \left (\ \dfrac{b}{2a} \right )^2 + b \left (\ \dfrac{b}{2a} \right ) + c = \dfrac{\cancel ab^2}{4a^{\cancel 2}}  \dfrac{b^2}{2a} + c = \dfrac{4ac}{4a} + \dfrac{b^2}{4a}  \dfrac{2b^2}{4a} = \dfrac{4ac  b^2}{4a}.$ And you are told in this case that maximum is $4 = \dfrac{4ac  b^2}{4a} = \dfrac{4ac  (2a)^2}{4a} = \dfrac{4ac  4a^2}{4a} = c  a.$ You have one equation with two unknowns, and you certainly can find a and b in terms of c. If you are supposed to find numeric answers, then you need a third equation. Are you told something else about this parabola? 
November 13th, 2016, 08:02 AM  #3 
Newbie Joined: Sep 2016 From: zambia Posts: 27 Thanks: 0 
thanks very much for the help.......

November 13th, 2016, 09:16 AM  #4 
Senior Member Joined: Feb 2014 Posts: 112 Thanks: 1  

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