Find an equation of a line tangent to $y=x^4-6x$ and perpendicular to the line $x-2y+6 = 0$. How to solve?

Question

This is all I got at the moment:

$y=x^4-6x$

$y'=4x^3-6$

$x-2y+6=0$

$y=x/2+3$

Answer 1

Let be $f(x)=x^4-6x$. To obtain a generic tangent line passing through a point $(t,f(t))$ take a generic line $y=mx+q$ and impose $m = f'(t)$ and the passage through the point. You have: \begin{gather} y=f'(t)(x-t) + f(t) \\ y=(4t^3-6)(x-t) + t^4-6t \end{gather}

Now you have to find a value of $t$ such that the angular coefficient of the line $y$ is $\dfrac{-1}{\frac{1}{2}}=-2$ (perpendicularity condition). You have ($t\in \mathbb{R}$): $$4t^3 - 6 = -2 \quad\Longrightarrow \quad t=1$$ Then $$ y=(4-6)(x-1)+1-6 = -2x-3. $$

Answer 2

Hint:

So, the gradient of the tangent has to be $$=\dfrac{-1}{\dfrac12}=-2$$

$$\implies4x^3-6=-2\implies x=?,y=x^4-6x=?$$