Question

If the tangent to the graph of the function `y = (x- 1)/x` passes through the point `(4, 1)`, then what is the equation of the tangent line?

Answer 1

The equation is `y = 1/4x`

Explanation:

We want the graph to pass through `(4, 1)`. The slope of the line is given by the derivative of `f(x)`.

`f(x) = 1 - 1/x`

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

Now

`y - y_1 = m(x -x_1)`

`y - 1 = 1/x^2(x - 4)`

`y = 1/x - 4/x^2 + 1`

We will need a second equation. Because the point `(x, y)` must also lie on the graph of `f(x)`, our second equation will simply be

`y = 1 - 1/x`.

We can immediately solve through substitution.

`1 - 1/x = 1/x - 4/x^2 +1`

`0 = 2/x - 4/x^2`

`0 = (2x - 4)/x^2`

`0 = 2x - 4`

`4 = 2x`

`x = 2`

Therefore, the slope of the line will be `f'(2) = 1/2^2 = 1/4`.

`y - 1 = 1/4(x - 4)`

`y - 1 = 1/4x - 1`

`y = 1/4x`

Let's confirm graphically.

The point in orange is the point of tangency, the point in green is the point `(4, 1)`, the line in red is the tangent line and the curve in blue is the function `f(x)`.

This proves that our answer is viable.

Hopefully this helps!