Differential Calculus (help pls)

Question

If y= ax^3 + bx^2 and its point of inflection is at (2, 4), then a=

Answer 1

a =-1

point of inflection is where the 2nd derivative =0

(2,4)

y=ax^3 +bx^2

y' =3ax^2 +2bx

y" =6ax +2b=0

divide by 2

3a +b = 0

plug the inflection point into the original equation, x=2, y=4

y=ax^3 +bx^2

4 =a(2^3) +b(2^2)

4= 8a +4b

divide by 4

1 =2a +b

0 =3a +b

subtract

3a-2a = 0-1

a =-1