Question

Find an equation of the tangent line to the graph at the given Point
f(x)=6/x^2+2
(2,1)

Answer 1

Answer:

Step-by-step explanation:

$f(x)=\frac{6}{x^2+2},\quad (x_0,y_0)=(2,1)\\f'(x)=\frac{-12x}{(x^2+2)^2}\\y-y_0=f'(x_0)(x-x_0)\\y-1=f'(2)(x-2)=\frac{-12\cdot 2}{(2^2+2)^2}(x-2)=-\frac{2}{3}(x-2)\\y=-\frac{2}{3}x+\frac{7}{3}$