2016/012) Suppose that $f(x+3)=3x^2+7x+4\cdots(1) $ and $f(x)=ax^2+bx+c\cdots(2)$ What is $a+b+c$

we have from 2nd equation $f(1) = a + b + c\cdots(3)$
Now we need to evaluate f(1)
as $f(x+3) = 3x^2 + 7x + 4$
so $f(x) = 3(x-3)^2 + 7(x-3)  + 4\cdots(4)$
putting x = 1 we get $f(1) = 3 * (-2)^2 + 7 * (-2) + 4 = 2$
so $a + b+ c = 2$