Finding the equation parallel to a line segment in my Calculus assignment, one of the questions ask to find the equation of a straight line that runs parallel to a line segment $PQ$, where $P(1,-2)$  & $Q(4,16)$, and touches the graph of $f(x)=x^3 - 3x^2$ at one point only given it has a negative y-intercept.
I honestly think this question is impossible and the best I've managed to do is find $PQ$ which is equal to $y=6x-8$, which I'll denote as $g(x)$.
I've tried to make $f'(x)$ equal to the gradient of $PQ$ but ended up with an answer that looked entirely wrong.
Prior to the question, I was asked to find point $Q$ when only given $P(1,-2)$, the $x$ point of $Q$ as $4$ and a gradient of $PQ$ as $6$. I simply used the gradient formula to obtain the coordinates of $Q$ as $(4, 16)$.
Then I substituted the point $P$ into $y=mx+c$ to obtain $c$ which gave me $g(x)=6x-8$, the equation for $PQ$
I then differentiated $f(x)$ to obtain $f'(x)=3x^3 - 6x$ and made the derivative equal to the gradient of $g(x)$ which gave me the quadratic function $3x^2 - 6x - 6$ which I then factorised  to obtain $x = 1±√3$.
I then substitute both values of $x$ into $3x^2 - 6x - 6$ but got lost in the middle of the process.
Any help would be much appreciated, thanks!
 A: The equation of the line is correct: $r : y = 6x - 8$
The derivative of the function is: $f'(x) = 3x^2-6x$
If we want the intersection point between the curve and the line, we clearly equate them:
$$6x-8 = 3x^2-6x$$
That is
$$3x^2 - 12x + 8 = 0 \longrightarrow x = \dfrac{1}{3}\left(6 \pm 2\sqrt{3}\right)$$
There are two solutions for $x$. One of them will provide you to find the tangent point $T = (x, y)$ with $y < 0$, considering that $T\in r$.
Using $x =\dfrac{1}{3}\left(6 - 2\sqrt{3}\right) $ and substituting it into the equation of $r$ you get
$$y = 2 \left(6-2 \sqrt{3}\right)-8$$
which is negative indeed.
A: We are looking for the intersection between a line of slope $6$ and real intercept $d\in \mathbb{R}$ and $f(x)$. The line is hence described by:
$$l(x)=6x+d, \text{for $d\in \mathbb{R}$}.$$
We are essentially looking for conditions for the polynomial given by:
$$p(x)=f(x)-l(x)=x^3-3x^2-6x-d$$
to have exactly one real root. Such condition is guaranteed by requiring that the discriminant $\Delta_3(d)$ of $p(x)$ satisfies:
$$\Delta_3(d)<0.$$
Do you know how to compute $\Delta_3$ and solve the above inequality in terms of $b$?
