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How do you minimize this?

  1. Jul 13, 2006 #1
    [tex]aX^2 + bX + c[/tex]

    where X belongs to the reals.

    I am reading an analysis book and one of the problems asks me to minimize this general form polynomial.

    I know that normally you would take the derivative and set it equal to zero, but in this case does that make sense? I would get the minimum is b when X = 0.

    If that is the answer, how does that help me to show that

    [tex] b^2 -4ac \leq 0 [/tex]

    Thank-you for the help.
  2. jcsd
  3. Jul 13, 2006 #2
    Are you sure about that? What is the derivative of that function? Also are there any conditions on a, b, and c because I don't think you can find a minimum without conditions, and in fact the function may not even have a minimum without them.
  4. Jul 13, 2006 #3


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    First, if you really mean the general [itex]ax^2+ bx+ c[/itex] where a,b,c can be any numbers, then it may not have a minimum! It will have a minimum if and only if a> 0.

    You don't really need calculus for this: complete the square!
    If [itex]y= ax^2+ bx+ c[/itex] then y= a(x^2+ \frac{b}{a}x)+ c. We can make the part inside the parentheses a perfect square by adding [itex]\left(\frac{b}{2a}\right)^2= \frac{b^2}{4a^2}[/itex]. Of course, we also need to subtract it.
    [tex]y= a(x^2+\frac{b}{a}x+ \frac{b^2}{4a^2}-\frac{b^2}{4a^2})+ c[tex]
    [tex] = a(x^2+\frac{b}{a}x+ \frac{b^2}{4a^2})- \frac{b^2}{4a}+ c[/tex]
    [tex] = a(x+ \frac{b}{2a})^2+ c- \frac{b^2}{4a}[/tex]
    Since a square is never negative, if a> 0, the minimum value occurs when the squared term is 0. That is, when [itex]x= -\frac{b}{2a}[/itex] and, in that case, [itex]y= c-\frac{b^2}{4a}[/itex]. If a< 0, then that is a minimum.

    As for
    I don't know how in the world you got that result. Did you take the derivative and then set X= 0?? Set the whole derivative equal to 0 and solve for x. The derivative of [itex]y= ax^2+ bx+ c[/itex] is [itex]y'= 2ax+ b[/itex]. That equals 0 when [itex]x= -\frac{b}{2a}[/itex] just as above. And in that case, [itex]y= \frac{b^2}{4a}- \frac{b^2}{2a}+ c= c- \frac{b^2}{4a}[/itex] again, just as before.

    It doesn't! If a, b, c can be anything in the general formula then [itex]b^2- 4ac[/itex] can be anything. Of course, that is the discriminant in the quadratic formula. If it is negative, then the equation [itex]ax^2+ bx+ c= 0[/itex] has no real solutions so the graph does not cross the x-axis. Assuming that a> 0 so this has a minimum, the if the minimum is not negative, that is if [itex]c-\frac{b^2}{4a}\ge 0[/itex] then, geometrically the graph clearly does not cross the x axis (but may be tangent to it) so there are no real roots (except possibly the vertex itself). Algebraically, it is easy to see that if [itex]c- \frac{b^2}{4a}\ge 0[/itex] then, multiplying by the positive number 4a, [itex]4ac- b^2\ge 0[/itex] so [itex]b^2- 4ac\le 0[/itex].
    Last edited: Jul 13, 2006
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