Find an equation of the tangent line to a graph with given point

Question

Q) Find the equation of the tangent line to the graph at the given point.

f(x)=(8)/(x^2+4)

point=(-2,1)

Would greatly appreciate some guidance on this, I know we have to use the point-slope equation but I can't seem to get the right answer.

Thank you in advance for any help!

Answer 1

take the derivative to find the slope. derivative of u/v = vu' - uv' all over v^2. In this problem u=8, v=x^2+4, u'=0, v'=2x

f'(x) = -16x/(x^2+4)^2

plug in x=-2

f'(-2) = -16(-2)/((-2)^2 + 4)^2 = 32/8 =4 = slope

y=mx + b where m=4

y = 4x + b, plug in (-2,1) to solve for b

1 =-8 +b

b =9

y=4x +9 is the equation of the tangent line at the point (-2,1)