I have a function $f(x) = x^3 + ax^2 + bx+ c$ and I need to solve for the numbers $a$, $b$, and $c$. The numbers need to satisfy the following condition:
The slope of the secant line defined by points $P = (0, 1)$ and $Q = (1, 0)$, both on the graph of $f$, is the slope of the tangent line of $f$ at $x = 1 + \frac1{\sqrt{3}}$.
The wording of this question has me confused on how to find the solution. Any help would be great!
$$f(x)=x^3+ax^2+bx+c$$ $$P=(0,1)\Rightarrow f(0)=1\Rightarrow c=1$$ $$Q=(1,0)\Rightarrow f(1)=0\Rightarrow 1+a+b+c=0\Rightarrow a+b=-2$$ now we are told that the secant line passes through both of these points. We know: $$f'(x)=3x^2+2ax+b$$ $$\text{grad secant}=\frac{0-1}{1-0}=-1$$ and we are told that $$\text{grad secant}=\text{grad tangent}$$ also we know by definition that: $$\text{grad tangent}\times f'=-1$$ and so we get: $$f'\left(1+\frac1{\sqrt3}\right)=1$$ going back to our equation from before: $$1=f'\left(1+\frac1{\sqrt3}\right)=3\left(1+\frac1{\sqrt3}\right)^2+2a\left(1+\frac1{\sqrt3}\right)+b$$ now we have a system of equations: $$\begin{pmatrix}2a\left(1+\frac1{\sqrt3}\right)&1\\1&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}1-3\left(1+\frac1{\sqrt3}\right)^2\\-2\end{pmatrix}$$ which can be solved for $a,b$ and we already know $c$