Find the equation of the line tangent to the point.

Question

Hello, I have been working on this problem and I have my work attached. I thought I got the right answer, but when I checked my answer I found that point P was not an inflection point of the equation after I found values of b and c. I don’t know where my mistake was so any help would be appreciated!

Answer 1

Your solution looks correct to me, so you must have made some error in checking the solution. Differentiating $x^3+3x^2+3$ twice produces $6x+6$, which clearly vanishes at $x=-1$, just as you have in the lower-left corner.

Perhaps you’re thinking that the graph must “flatten out” at this point for it to be an inflection point, but all that’s required is that the curve go from being concave to convex, or vice-versa.

Answer 2

Since we have $$y'=3x^2+2bx$$ $$y''=6x+2b$$ we get $$y''=0$$ if $$x=-\frac{b}{3}$$ since $$P(-1,5)$$ we get $$b=3$$ and $$5=-1+27+c$$ so $$c=-21$$