Question

How do you find the equation of the tangent line to the graph of the given function `f(x)= x/(x-2)`; at (3,3)?

Answer 1

`y = -2x + 9`

Explanation:

To find the equation of the line tangent to the graph of the function at a given point, in your case`(3,3)`, you need to calculate the first derivative of the function first.

The first derivative of the function will give you the slope of this line

`color(blue)("slope" = m = d/dxf(x))`

Use the quotient rule to differentiate the function

`d/dxf(x) = ([d/dx(x)] * (x-2) - x * d/dx(x-2))/(x-2)^2`

`f^'(x) = [1 * (x-2) - x * 1]/(x-2)^2`

`f^'(x) = (color(red)(cancel(color(black)(x))) - 2 - color(red)(cancel(color(black)(x))))/(x-2)^2 = -2/(x-2)^2`

The slope of the tangent line at point `(3,3)` will be

`m = f^'(3) = -2/(3-2)^2 = -2/1 = -2`

The equation of the line in point slope form will be

`color(blue)(y - y_1 = m * (x - x_1))`

`y - 3 = -2 * (x-3)`

`y = -2x + 6 + 3`

`y = -2x + 9`