All I've been out of school for some time and I'm looking for some help as I'm reviewing to prepare to go back to school. In my old notes, I found this question and I have no idea how to solve it.
For what values of $a$ and $b$ is the line $-4x + y = b$ tangent to the curve $y = ax^3$ when $x = -2$?
Because we want $y=4x+b$ is tangent to the curve $y=ax^3$ at $x=-2$ then $ \frac{dy}{dx} $ at $x=-2$ must be equal to the gradient of $y=4x+b$ which is $4$. So $$ 4=\frac{dy}{dx}\Bigg |_{x=-2}= 3ax^2 = 3a(-2)^2= 12a \implies a= 4/12 = 1/3$$ To find $b$ we must know the value of $y$ when $x=-2$. The value $y$ must be $y= ax^3=-8/3$. Because the line $y=4x+b$ also pass $(-2,-8/3)$ (because its tangent to it) then by substitution $(-2,-8/3)$ to $y=4x+b$ we have $b=16/3$.
Good luck Joshua !