We find the equation of the tangent line to $f(x) = x^3-x$ at the point $(k, k^3-k)$.
First, the derivative gives the slope
$$f'(x) = 3x^2 -1$$
So we have a line with slope $3k^2-1$ that goes through the point $(k,k^3-k)$. What is its equation? We use the point-slope form
$$y-(k^3-k) = (3k^2-1)(x-k)$$
$$y-k^3 +k = 3k^2x-x-3k^3+k$$
$$y+2k^3 = 3k^2x-x$$
We want to know which values of $k$ make this line pass through $(x,y)=(-2,2)$, so we plug in those values:
$$2+2k^3=3k^2(-2)-(-2)$$
$$2k^3 = -6k^2$$
$$k^3 = -3k^2$$
$$k=0 \,\,\,\,\text{ or }\,\,\, k=-3$$
This means that there are two tangent lines that work, one at $(0,0)$, and the other at $(-3,-24)$. You should be able to find the equations of these lines easily if necessary.
what about equaly problem by my function look like this: $x^2-xy+y^2-1=0$. My question is how to solve equaly problem but my function is implicitly and I can't easily solve for $y$
You want to find the number of tangents to a curve $x^2-xy+y^2-1=0$ that all pass through the origin.
Let $(s,t)$ be the tangent point on the curve, i.e. $$s^2-st+t^2-1=0\tag1$$
We have $$2x-y-xy'+2yy'=0\iff (-x+2y)y'=-2x+y$$
Suppose that $-x+2y=0$, then $-2x+y=0$. Hence, $x=y=0$ but this point is not on the curve. Hence, $-x+2y\not=0$.
So a line tangent at $(s,t)$ is given by
$$y-t=\frac{-2s+t}{-s+2t}(x-s)$$
Setting $x=y=0$ gives
$$0-t=\frac{-2s+t}{-s+2t}(0-s),$$
i.e.
$$s^2+t^2-st=0\tag2$$
However, there is no $(s,t)$ satisfying both $(1)$ and $(2)$. Hence, there is no such tangent.
In general, you would have two equations on two variables.
Best Answer
The parametric form is $t,t^3+t^2-22t+20$
The gradient at $x=t$ will be $$3t^2+2t-22$$
So, the equation of the tangent at $x=t$ will be $$\frac{y-(t^3+t^2-22t+20)}{x-t}=3t^2+2t-22$$
This has to pass through the origin $(0,0)$
Hope you can take it home from here