Determine values of a,b,c such that the graph $\displaystyle y=ax^2+bx+c$ has a relative maximum at $\displaystyle (3,12)$ and crosses the y axis at (0,1)
first, you have two points on the curve given to you ...
(0,1) ... $\displaystyle 1 = a(0^2) + b(0) + c$ ... now you know $\displaystyle c = 1$
(3,12) ... $\displaystyle 12 = a(3^2) + b(3) + 1$
this gives you the equation $\displaystyle 9a + 3b = 12$ , or simplified, $\displaystyle 3a + b = 4$
now you need another equation ... relative extrema occur at critical values; in this case where $\displaystyle y' = 0$.
$\displaystyle y = ax^2 + bx + 1$
$\displaystyle y' = 2ax + b$
$\displaystyle 2ax + b = 0$ at $\displaystyle x = 3$ ...
$\displaystyle 6a + b = 0$
there is your second equation in terms of $\displaystyle a$ and $\displaystyle b$ ... solve the system.