The graphs of the functions $f(x)=x^3+(a+b)x^2+3x−4$ and $g(x)=(x−3)^3+1$ touch. Express a in terms of b.
The solution in the textbook is $a=−\frac{(27+11b)}{11}$
I've tried looking for a worked solution, but couldn't find anything. Help would really be appreciated, thanks.
Set the functions equal to each other and expand to get $$x^3-9x^2+27x-27+1=(a+b)x^2+3x-4,$$ and this simplifies to give you $$(a+b+9)x^2-24x+22=0.$$ If we require the functions to intersect in just one point, then the equation $$24^3-88(a+b+9)=0$$ must be true (by setting the discriminant to $0$), which when you solve for $a$ gives $$a=-\frac{27}{11}-b,$$ as desired.