Given point and tangent line The function is $f(x)=1/x^2$. 
I have to find equation of the tangent line that also goes through the point $(0,12)$.
I know how to construct this problem in desmos: https://www.desmos.com/calculator/bj1as3oqo0
So by reading the graph I know solutions are: 


*

*$y=-16x+12$

*$y=16x+12$


But how do I find this equations only by calculation? How do I figure out when the line becomes the tangent in this case?
I think I'm pretty good at understanding derivatives and tangent lines, but this one is a hard case for me. 
 A: $$f(x)=\frac1{x^2}=x^{-2}\implies f'(x)=-2x^{-3}$$
The parametric equation of $\displaystyle f(x)=\frac1{x^2}$ is $\displaystyle\left(t,\dfrac1{t^2}\right)$
$\displaystyle\implies  f'(x)_{\text{ at }\left(t,\dfrac1{t^2}\right)}=-\frac2{t^3}$
So, the equation of the tangent will be $$\frac{y-\dfrac1{t^2}}{x-t}=-\frac2{t^3}\iff 2x+yt^3=3t$$
Now use the fact : this line passes through $(0,12)$ to find $t$
A: We have, the equation of any tangent at $x_0$ is: $g(x)=f'(x_0)(x-x_0)+f(x_0)$, $(x_0 \neq 0)$.
We have to find the tangent such as $g(0)=12$ which means $f'(x_0)(-x_0)+f(x_0)=12$ so, $\frac{2}{x_0²}+\frac{1}{x_0²}=12$ By multiplicating both sides by $x_0²$ we find: $3=12x_0²$ Wich means either $x_0=1/2$ or $x_0=-1/2$ .
You can verify by replacing $x_0$ by 1/2 or -1/2 in the equation of the tangent, you will find $y=-16x+12$ and $y=16x+12$
A: Hint: (a naive approach, but possibly the only one) Take the derivative at any point $x_0$ to determine the $y$-intercept of the tangent line at $x_0$. The $y$-intercepts are a function of $x_0$; find the points where that function is equal to $12$.
[Fair warning: I haven't gone through the calculations, so maybe you end up with some nasty polynomial. But without thinking too hard, looks like it should be a quadratic equation in the end.]
