Minimum value of quadratic equation I know that $ax^2+bx+c$ if $a < 0$ it has maximum value and we find $m = -b/2a$
but if $a>0$ it should have minimum value how can find it?
Thanks in advance.
 A: It's the same $m = \frac{-b}{2a}$. One way you can notice this is as follows: Suppose the 2nd degree polynomial $ax^2+bx+c$ has $a>0$. Then multiply through by $-1$ and you'll have a 2nd degree polynomial with $a<0$. As you pointed out, the second polynomial will have minimum at $x = \frac{-b}{2a}$. This must mean the same value of $x$ gives a maximum in the first polynomial. 
A: For a more technical and deeper answer , you might want to use basic Calculus . 
If you have studied some basic derivative , we know that the local maxima or local minima or the function occurs when the derivative of the of the function is $0 .$
This is true because derivative , in a sense means to figure out the slope of point in a function . And since the the function is at  minima/maxima , the slope should be $0$. $($Compare this with the time when , just before the fall , you are at the top of the roller coaster and you seem to be still at the max height$.)$
Now using some basic calculus , we find the derivative of $ax^2+bx+c$ is $2ax+b$ . Since maxima/minima occurs at $0$ , it follows that:
$$2ax+b=0\implies x=-\dfrac b{2a}$$
To figure out the whether this is a local minima/maxima , we differentiate it again to get $2a$. If $2a\gt0$ , the function is at the minima . Else if $2a\lt0$ , the function is at it's maxima.
Clearly if $a\gt 0$ , then $x=-\dfrac b{2a}$ gives us the minima , else it gives us the maxima of the function.

In case you are still confused , behold this :


Source : Desmos

