Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 November 13th, 2016, 06:42 AM #1 Newbie   Joined: Sep 2016 From: zambia Posts: 26 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: 430 Thanks: 173 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? Thanks from life24
 November 13th, 2016, 08:02 AM #3 Newbie   Joined: Sep 2016 From: zambia Posts: 26 Thanks: 0 thanks very much for the help.......
November 13th, 2016, 09:16 AM   #4
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 Originally Posted by jeho thanks very much for the help.......
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