The given are
Two x-intercepts
y-intercept(0,-4)
Maximum at (2,4)
i tried everything i know...its been a long time since I have been doing math problems but the only way i thought about was to use -b/2A but i can't go anywhere after that...if possible just a hint to solve it is ...or if possible anyone tell me how to solve it?? The only other way i can think is point slope formula....
You can start by writing out the general form of a quadratic polynomial: $P(x)=ax^2+bx+c$. Saying that the y-intercept is $(0,-4)$ is the same as saying that $P(0)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be $P'(x)=2ax+b$. Since the derivative of a differentiable function at a maximum is 0, you must have $P'(2)=0$ or $4a=-b$. Since the point $(2,4)$ is assumed to be in the graph of the function, we know that $P(2)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have
$$4a=-b \;\;\;\;\text {and}\;\;\;\ 4a=8-2b$$
Which gives $b=8$ and $a=-2$, so we get $P(x)=-2x^2+8x-4$
EDIT
Another way to do it, as pjs36 said, would be to use the fact that every quadratic equation is of the form $P(x)=\alpha(x-\beta)^2 +\gamma$ for some $\alpha,\; \beta,\; \gamma$. When the equation is written in this form it becomes clear that $\alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-\beta =0$. Since the maximum is at $x=2$, we have $\beta =2$. Also, the value of $P$ at the maximum will be $4=P(2)=\alpha (0)^2 + \gamma = \gamma$, so $\gamma =4$. Lastly, we know that the y-intercept is $(0,-4)$, so that $-4=P(0)=\alpha(0-\beta )^2 + \gamma = 4\alpha +4$, which gives $\alpha=-2$. Now we can expand the original expression and we get $P(x)=\alpha (x- \beta)^2 +\gamma =-2(x-2)^2 + 4=-2x^2 +8x -4$, so we get the same result as before.