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.