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.
[tex]y= a(x^2+\frac{b}{a}x+ \frac{b^2}{4a^2}-\frac{b^2}{4a^2})+ c[tex]
$$= a(x^2+\frac{b}{a}x+ \frac{b^2}{4a^2})- \frac{b^2}{4a}+ c$$
$$= a(x+ \frac{b}{2a})^2+ c- \frac{b^2}{4a}$$
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.

As for

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.

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 $y= ax^2+ bx+ c$ 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.

If that is the answer, how does that help me to show that
$$b^2- 4ac\le 0$$

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 if the minimum is not negative, 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 $b^2- 4ac\le 0$.