so let's say the equation is y = ax² + bx + c, where a,b,c are constants.
we know that f(3) = 5, or namely when x = 3 , y = 5.
we also know that f'(3) = 6, or namely the slope at the point (3,5) is 6.
we also know that at (3,5) the 2nd derivative is 4, so a positive number simply tell us its concavity, is up, so is a parabola opening upwards.
![y=ax^2+bx+c\implies \left. \cfrac{dy}{dx}=2ax+b \right|_(x=3)~~ = ~~ 6\implies 6=2ax+b \\\\\\ 6=2a(\stackrel{x}{3})+b\implies 6=6a+b\implies 6-6a=b\implies \boxed{6(1-a)=b} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{d^2y}{dx^2}=2a \right|_(x=3)~~ = ~~4\implies 2a=4\implies a=\cfrac{4}{2}\implies \boxed{a=2}~\hfill \boxed{-6=b} \\\\[-0.35em] ~\dotfill](https://img.qammunity.org/2023/formulas/mathematics/college/4y5nasyc4s7mez6wme5afc51kyws4ektlk.png)
