How Does Completing the Square Help Minimize a Quadratic Function?

[wooby]

aX^2 + bX + c

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

b^2 -4ac \leq 0

Thank-you for the help.

 

[d_leet]

aX^2 + bX + c

where X belongs to the reals.


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.

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.

 

[HallsofIvy]

aX^2 + bX + c

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

b^2 -4ac \leq 0

Thank-you for the help.

First, if you really mean the general ax^2+ bx+ c 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 y= ax^2+ bx+ c then y= a(x^2+ \frac{b}{a}x)+ c. We can make the part inside the parentheses a perfect square by adding \left(\frac{b}{2a}\right)^2= \frac{b^2}{4a^2}. Of course, we also need to subtract it.
y= a(x^2+\frac{b}{a}x+ \frac{b^2}{4a^2}-\frac{b^2}{4a^2})+ c<br /> = a(x^2+\frac{b}{a}x+ \frac{b^2}{4a^2})- \frac{b^2}{4a}+ c<br /> = a(x+ \frac{b}{2a})^2+ c- \frac{b^2}{4a}<br /> Since a square is never negative, if a> 0, the minimum value occurs when the squared term is 0. That is, when x= -\frac{b}{2a} and, in that case, y= c-\frac{b^2}{4a}. If a< 0, then that is a minimum.<br /> <br /> As for <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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. </div> </div> </blockquote> 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 <b>solve</b> for x. The derivative of y= ax^2+ bx+ c is y'= 2ax+ b. That equals 0 when x= -\frac{b}{2a} just as above. And in that case, y= \frac{b^2}{4a}- \frac{b^2}{2a}+ c= c- \frac{b^2}{4a} again, just as before.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> If that is the answer, how does that help me to show that <br /> b^2- 4ac\le 0 </div> </div> </blockquote> It doesn't! If a, b, c can be anything in the general formula then b^2- 4ac can be anything. Of course, that is the discriminant in the quadratic formula. If it is negative, then the equation ax^2+ bx+ c= 0 has no real solutions so the graph does not cross the x-axis. Assuming that a> 0 so this has a minimum, the <b>if the minimum is not negative</b>, that is if c-\frac{b^2}{4a}\ge 0 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 c- \frac{b^2}{4a}\ge 0 then, multiplying by the positive number 4a, 4ac- b^2\ge 0 so b^2- 4ac\le 0.