Find the equation of the tangent to the function f(x) = -2/x, at the point when x = 2.

Question

Find the equation of the tangent to the function f(x) = -2/x, at the point when x = 2.

Answer 1

f'(x) = (-2/x)' = (-2x^-1)' = 2x^-2 = 2/x^2

f'(2) = 2/2^2 = 2/4 = 1/2

f(2) = -2/2 = -1

y= (1/2)x + b

-1 = (1/2)(2) + b

-1 = 1 + b

b =-2

y= x/2 - 2 is the equation of the line tangent to the given hyperbolic equation y= -2/x

or

2y = x - 4 to eliminate fractions

y=-2/x is a hyperbola with 2 branches, on in quadrant IV and II with asymptotes x=0 and y=0

the tangent line at x=0 touches the hyperbola in quadrant IV at (2,-1) with a slope of 1/2, y intercept -2 and x intercept 4

Answer 2

First find f '(x). Then find f '(2). This is the derivative of the function f at the point x = 2. f '(2) is also the slope of the tangent line at x = 2.

A line requires a slope and a point through the line, so we just need the point. The tangent line intersects f(x) at x = 2. You can find the point that lies on both the tangent line and the function by plugging 2 into f(x). Your point will be (2, f(2)).

Lastly, find the equation of the line using the point and the slope.