Verified answer

f '(x) = -38/x^3

f '(2) = -38/2^3 = -38/8 = -19/4

Tangent line's equation is

y - 19/4 = -19/4(x - 2)

y - 19/4 = -19x/4 + 19/2

y = -19x/4 + 57/4

Begin by checking the given point is on the curve:

f(x) = 19/x² = 19x^-2

19/4 = 19/4

Next, take the first derivative.

f'(x) = -38x^-3

Evaluate at x = 2, to obtain the slope (m) of the line.

m = -38/8 = -19/4

Using the slope-intercept form of a line, y = mx + b, set y = 19/4, x = 2, m = -19/4 and solve for b.

19/4 = (-19/4)2 + b

b = 57/4

The equation of the line is:

y = (-19/4)x + 57/4

Standard form:

19x + 4y = 57

First, find the slope by generating the derivative

f(x) = 19 / x^2

f'(x) = -2 * 19 / x^3 = -38 / x^3

f'(2) = -38 / 2^3 = -19 / 4

So we need a line with a slope of -19/4 that passes through (2 , 19/4)

y - 19/4 = (-19/4) * (x - 2)

4y - 19 = -19 * (x - 2)

4y - 19 = -19x + 38

4y = -19x + 57

4y = (-19) * (x - 3)

y = (-19/4) * (x - 3)