If f(x)=ax^2+b
f(-3)=-3 and f'(-3)=2
find the coefficients a and b.
f(x) = ax^2 + b
f ' (x) = 2ax
given :
f ' (-3) = 2 ----> 2a(-3) = 2
-3a = 1
a = -1/3 , so f(x) = (-1/3)x^2 + b
f(-3) = -3
---> (-1/3)(-3)^2 + b = -3
-3 + b = -3
b = 0
so a = -1/3, b = 0 and f(x) = (-1/3)x^2
check:
f(-3) = -(1/3)(9) = -3
f ' (x) = -(2/3)x
f '(-3) = -(2/3)(-3) = 2
all looks good.