The graph of f(x) =ax^3+bx^2 has a local turning point at (2;-4) find the value of a and b

Question

The turning point is the only info given with the graphs equation

Answer 1

y=ax^3+bx^2...or y=x^2*(ax+b)

a "terning point" meen the slope chaenj from positiv tu negativ or the opposit

for a kontinyuus kerv, that meen it gotta go thru ZERO.

slope=dy/dx=3ax^2+2bx

d2y/dx2=6ax+2b=0 at x=2, so 2*6a=-2b, or b=-6a

yu sae kerv go thru point=(2,-4)...y=ax^3+bx^2...8a+4b=-4...4b=-8a-4...b=-2a-4

thus b=-6a=-2a-4, so 2a-6a=-4...-4a=-4...a=1

b=-6a=-6