Question

For what values of a and b is the line 3x+y=b tangent to the curve y=ax^3 when x=–3?

Answer 1

Differentiate y=ax^3 gives:dy/dx=3ax^2 from 3x+y=b rearranging y=-3x+b m=-3 therefore 3ax^2=-3 ax^2=-1 since x=-3 a(-3)^2=-1 9a=-1 a=-1/9 (b)when x=-3 y=0 0=-3(-3)+b b=-9