How to find the derivative of a curve where a tangent line has a slope of -4?

Question

I'm having a little problem with my math homework. I searched on the internet but came up even more confused.

So here's the problem:

If $y = \frac{1}{x}$, find $y'$ and use this result to find the points on the curve $y = \frac{1}{x}$ where the tangent line has the slope $-4$.

I found the derivative of the function which is $y' = -\frac{1}{x^2}$. However, I do not know how to proceed.

Answer 1

you must solve the equation $$\frac{-1}{x^2}=-4$$ for $x$

Answer 2

The slope of a curve is its derivative. So you have $$y=x^{-1}\implies y'=-x^{-2}=-4\implies \frac1{x^2}=4\implies x^2=\frac14\implies x=\pm\frac12$$ Substituting these values into $y$ gives the two pairs of solutions $$(x,y)=\left(\pm\frac12, \pm2\right)$$