Find the sign of $a,b,c$ in $ax^2+bx+c$ given the graph and a coordinate on it. 

So my first approach was that, we see that there are $2$ roots. And one is negative and one is positive. $a$ would be evidently positive. The positive one's modulus is bigger than the negative one's. So the sum of roots would be $+ve$, that is $\frac{-b}a$. So if $a$ is positive and  sum of roots is positive, then $b$ has to be negative. And similarly, product of the roots will be $-ve$, that is $\frac ca$, so $c$ will be negative. Which gives the option $b)$ as correct. 
But if we see, point $A$ will be the minimum point. And at the minimum point value of function is $\frac{-D}{4a}$. But generally the value is $\frac{-D^2}{4a}$ , because 
[Edit] let $y=ax^2+bx+c$. Now let $a$ be positive. So minimum will be at $x=\frac{-b}{2a}$. Substitute. $y=a(\frac{b^2}{4a^2})-\frac{-b^2}{2a}+c$ 
$y=\frac{b^2-2b^2+4ac}{4a}=\frac{-(b^2-4ac)}{4a}=\frac{-D^2}{4a}$
 This means that $b^2=4ac$. Which is completely different and unsupportive of the graph. So where did I do it wrong.
 A: a is positive and c is negative are pretty much clear by looking at the shape and y-intercept of the graph. b is negative since the x coordinate of the turning point is positive. As a is positive you need b to be a negative valune to change the sign.
A: Assuming that you understand some Calculus!
Let $f(x) = ax^2 + bx + c$ (a second degree polynomial in the single variable $x$) where $a,\ b,\ c$ and $x$ denote real numbers where $a \neq 0$.
Using the Calculus: We can see where some of the given information is coming from and then using it to determine the answer to the questions
$f^\prime(x) = 2ax + b = 0$ exactly when $x = -\frac{b}{2a}$ which is the $x$-coordinate of the point where the graph of $f(x)$ as a single turning point given as $A$. The corresponding $y$-coordinate of $A$ is $f(-\frac{b}{2a}) = \frac{4ab-b^2}{4a} = - \frac{b^2-4ac}{4a}$.  Hence the coordinates of the turning point (known as the vertex) is then $(-\frac{b}{2a},\ -\frac{b^2-4ac}{4a})$.  Hence, we can see that in the given problem $D\equiv b^2-4ac$ (known as the Discriminant and is commonly used to determine whether or not $f(x)$ has real roots---$D\geq 0$ tell us "Yes, real roots exists" and $D < 0$ tell us "No real roots exist".)
Recall the quadratic formula (the roots of $f(x)$) $$x\equiv -\frac{b}{2a}\ \pm\ \frac{\sqrt{D}}{2a} = -\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$$ Consider the sum and product of the roots!
Next, using the second derivative $f^{\prime\prime}(x)$ tell us the concavity (upwards or downwards) of the graph of $f(x)$.
$$f^{\prime\prime}(x) = (2ax + b)^{\prime} = 2a$$Now if $f^{\prime\prime}(x) > 0$ then the graph of $f(x)$ is concaved upwards and if $f^{\prime\prime}(x) < 0$, the graph of $f(x)$ is concaved downwards.
Now the $y$-intercept (given as $P$) is the point where the graph of $f(x)$ intersects the $y$-axis and has the form $(0,\ f(0))$ (or $(0,\  c)$).
Let's determine the sign of the coefficents $a,\ b$ and $c$ of $f(x)$ from the given graph.

*

*The given graph is concaved upwards, so $f^{\prime\prime}(x) = 2a > 0$ exactly when $a > 0$; that is, the coefficient $a$ is positive.


*The turning point given as $A$ lies in the fourth quadrant of the coordinate system and, so the point $A$ has a positive $x$-coordinate and a negative $y$-coordinate; that is,
$$-\frac{b}{2a} > 0\ \ \ \hbox{and}\ \ -\frac{b^2-4ac}{4a} < 0$$From the first inequality together with the fact that $2a > 0$ we easily deduce that $-b > 0$ or equivalently that $b < 0$; that is, the coefficent $b$ is negative.


*The $y$-intercept of the given graph lies directly below the origin $O$ and as such must have a negative $y$-coordinate. In this case, $f(0) = c < 0$ which means that the coefficient $c$ is negative.
In summary, this comes down to choice (B) in the question.
