Equation of the normal to a curve

Question

I am struggling to find the equation of the normal to the line: $$y = \frac{1}{x} - \frac{3}{x^2} - \frac{4}{x^3} + \frac{7}{4}$$ at $(-2,1)$. Any ideas would be appreciated. I believe I need to differentiate, but what do I do after that.

Answer 1

Tips:

• Remember $\;\biggl(\dfrac1{x^n}\biggr)'=-\dfrac{n}{x^{n+1}},\;$ so $$y'=-\frac2{x^2}+\frac{6}{x^3}+\frac{12}{x^4}. $$
• An equation of the straight line with given slope $m$, passing through a given point $(x_0,y_0)$ is given by the formula $$y-y_0=m(x-x_0).$$

The normal at the point with abscissa $-2$ will have slope $m=\dfrac{-1}{y'(-2)}$, and its equation will be $$y-1=m(x+2).$$